Algebra Chapter 7 Logarithm
Chapter 7 Logarithm Introduction
In the previous chapter, you studied the methods of solutions of the algebraic equations having indices.
In that cases, you have found how the values of x are determined by solving any equation of the type 2x = 8.
Now, it is not a matter Of job to you to say that the solution of 2X = 8 is x = 3,
Since 2x = 8
=> 2x = 23
=> x = 3.
Similarly, if 3x = 9, then x = 2, since 3x = 9
3x = 32
x = 2.
Thus it is easy to solve equations of this type.
But if the given equation is of type 2X = 11, i.e., 11 in the right-hand side cannot be easily expressed in the form of any simple index of 2, then it is undoubtedly very difficult to determine the value of x by the knowledge so far as attained by you.
To solve problems of this type, the mathematicians deducted a new method – a new chapter, the name of which is the logarithm.
It is now treated as a vast chapter of modern mathematics.
Read and Learn More WBBSE Solutions For Class 9 Maths
Chapter 7 Logarithm What Is A Logarithm
It can be simply said that a logarithm is a new type of expression of real numbers.
Such as, we can express 1 with the help of a logarithm as \(\log _2 2, \log _3 3, \log _4 4\)………… etc,.
\(1=\log _2 2=\log _3 3=\log _4 4=\)…………etc.
Similarly, 2 = \(\log _2 4,3=\log _2 8,4=\log _2 16,5=\log _2 25\) …… etc.
In this way, any real number can be expressed in the logarithmic form.
Definition of logarithm:
Let a and M be two real numbers and a > 0, a not = 1, and M > 0, then the real number x is called the logarithm of M with respect to the base a, if a = M.
Conversely, if ax = M where a > 0, a 1 and M > 0, then x = \(\log _a \mathrm{M}\).
Here \(\log _a \mathrm{M}\) is a unique real number when M ≠ 1, i.e., if \(\log _a \mathrm{M}=\log _b \mathrm{M}\), then a = b when M not = 1.
From the above definition, we can say that if any real number M can be expressed in the form at (where a > 0, a ≠ 1, and x is any real number), then its logarithm will be defined, otherwise not.
Now we have to see, which of the real numbers can not be expressed in the form at.
To do so, let M be a negative real number, i.e., M= -N for N > 0, then we want to know whether – N can be expressed in the form ax or not.
let ax = -N…………….(1)
If x < 0, then a-y = -N when x = -y for y > 0
⇒ \(\frac{1}{a^y}=-\mathrm{N} \Rightarrow 1=-\mathrm{N} \times a^y\)……………(2)
In (2) we have N > 0, a>0 and a ≠ 1, y > 0
∴ the value of the quantity (- N x ay ) can never be 1.
Also, if x = 0, then a° = – N
or, 1=- N
or, N = -1.
But N>0,
∴ it is impossible.
Further, if x > 0, then a > 0, since ax > 0 and a ≠ 1,
∴ ax ≠ – N.
Thus, if M< 0, then \(\log _a \mathrm{M}\) is undefined.
Hence logarithms of negative real numbers are undefined.
Again, if M = 0 we get ax = 0, which is impossible because a > 0, a ≠ 1, and x is a real numbers.
∴ if M= 0, then \(\log _a \mathrm{M}\) is undefined.
Hence logarithm of 0 is not defined.
Moreover, if a < 0 in M, then let a = -b for b > 0.
∴ ax = M⇒ (-b)x = M.
or, (-1)x.bx = M……(3)
Now, the value of (- 1)x in (3) will be positive if x is even and will be negative if x is odd.
∴ if x is odd and a < 0, then M < 0 which implies that \(\log _a \mathrm{M}\) is not defined (it is discussed earlier).
Also, if a = 0, then a = M
⇒ 0= M
or, M = 0 and we have already seen that if M = 0, then \(\log _a \mathrm{M}\) is undefined.
From all the above discussion we can now say that the definition “If ax = M, then x = log,M” is not true, i.e., is not defined in the cases-
1. If a < 0
2. If a = 0
3. If a = 1
4. If M < 0
5. If M = 0.
Now, we know that \(2^{-3}=\frac{1}{2^3}=\frac{1}{8}\)
∴ \(-3=\log _2\left(\frac{1}{8}\right)\)
Again, \(3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
∴ \(-2=\log _3 \frac{1}{9}\)
∴ any negative real number can be expressed in the logarithmic form, i.e., it is clear that the logarithm and the logarithmic form of any number are not the same.
However, ax = M is the expression of any real number in the form of an index, the index of which is x and we generally express this index x in the form of a logarithm.
Hence, any real number can be expressed in the form of a logarithm, but the logarithms of all real numbers are not defined.
Now, if ax = M, then we have, x =\(\log _a \mathrm{M}\).
∴ the logarithm of any real number must be with respect to some base.
Such as based on \(\log _a \mathrm{M}\) is a.
Also, if log, M is not uniquely defined, then it should be rejected.
Example-1.
∴ 4² = 16.
∴ 2 = \(\log _4 16\) conversely,
2 = \(\log _4 16\)
4² = 16.
Example-2.
∴ 25 = 32,
∴ 5 = \(\log _2(32)\) conversely,
5 = \(\log _2(32)\)
25 = 32.
Example-3.
∴ (-2)4 = 16
∴ \(\log _{(-2)} 16\),
Again, 24 = 16
4 = \(\log _2 16\).
∴ \(\log _{(-2)} 16=\log _2 16 \Rightarrow-2=2\), which is impossible.
∴ \(\log _{(-2)} 16\) is not defined.
Example-4.
If a = 0 in ax = M, then 0x = 0
⇒ x = \(\log _0 0\)
Again, 0² = 0
2 = \(\log _0 0 \text { and } 0^3=0 \Rightarrow 3=\log _0 0\)
∴ \(\log _0 0 \neq \log _0 0\), since 2 ≠ 3.
∴ If a = 0 in ax = M, then x = \(\log _a \mathrm{M}\)
Chapter 7 Logarithm Types Of Logarithm
Logarithms are of two types:
1. Common logarithms and
2. Natural logarithm.
1. Common logarithm:
Common logarithm was first formulated by Henry Briggs.
According to his name, this type of logarithm is also called the Briggarian method of the logarithm.
In this method, the base is assumed to be 10.
Therefore, the logarithm of any real number M (>0) with respect to the base 10 is called the common logarithm of that number.
For example,
\(10^2=100, ∴ 2=\log _{10} 100\)
\(10^3=1000, ∴ 3=\log _{10} 1000\)
\(10^{-1}=\frac{1}{10}, \quad ∴ -1=\log _{10}\left(\frac{1}{10}\right)\)…………etc.
Conversely, if \(\log _{10} 10\), then 10x = 10
x= 1,
∴ \(\log _{10} 10=1\).
If \(\log _{10} 100\), then 10x = 100
= 10²
x = 2.
2. Natural logarithm.
Natural logarithm was first formulated by John Napier.
According to his name, this type of logarithm is also known as the Naperian method of the logarithm.
In this method, a transcendental irrational number e is considered to be the base of the logarithm where
\(e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots \cdots \cdots \infty=2 \cdot 718281828 i.e., 2<e<3\)
Therefore, the logarithm of any real number M (M > 0) with respect to the base e is called the natural logarithm of that number.
For example, if ex = y, then x = \(\log _e y\), if ea = b, then a = \(\log _e b\)
Conversely, if \(\log _e 10=x \text {, then } e^x=10 \text {, if } \log _e y=z\), then ez = y.
Chapter 7 Logarithm Laws Of Logarithm
If M, N, a, and b are all positive real numbers, and if n is any real number then-
Formula 1: \(\log _a(M N)=\log _a M+\log _a N\)
Formula 2: \(\log _a\left(\frac{M}{N}\right)=\log _a M-\log _a N\)
Formula 3: \(\log _a M^n=n \log _a M\)
Formula 4: \(\log _a \mathbf{M}=\log _b \mathrm{M} \times \log _a b\)
Proof Of The Laws:
Formula 1: \(\log _a(M N)=\log _a M+\log _a N\)
Proof:
Let x = \(\log _a \mathrm{M}\)…………….(1)
and y = \(\log _a \mathrm{N}\)……………(2)
From (1) we get, ax = M (by definition), and from (2) we get, ay = N (by definition)
Now, \(a^x \cdot a^y=\mathrm{MN}\) [Multiplying (1) and (2) ]
or, \(a^{x+y}=\mathrm{MN}\)
∴ \(\log _a \mathrm{MN}=x+y \text { or, } \log _a \mathrm{MN}=\log _a \mathrm{M}+\log _a \mathrm{~N}\)
∴ \(\log _a(M N)=\log _a M+\log _a N\) (proved)
Corollary 1: If n is any positive integer, then
\(\log _a\left(\mathrm{M}_1 \mathrm{M}_2 \mathrm{M}_3 \cdots \cdots \mathrm{M}_n\right)=\log _a \mathrm{M}_1+\log _a \mathrm{M}_2+\log _a \mathrm{M}_3+\cdots \cdots \cdots+\log _a \mathrm{M}_n\)
For example: \(\log _2(2 \times 3 \times 4 \times 5 \times 6)=\log _2 2+\log _2 3+\log _2 4+\log _2 5+\log _2 6\)
Formula 2: \(\log _a\left(\frac{M}{N}\right)=\log _a M-\log _a N\)
Proof:
Let x= \(\log _a \mathrm{M}\)….… (1) and y = \(\log _a \mathrm{N}\)……(2)
From (1) we get, ax = M (by definition), and from (2)
we get, \(a^y=\text { N } ∴ \frac{a^x}{a^y}=\frac{\mathrm{M}}{\mathrm{N}} \text { or, } a^{x-y}=\frac{\mathrm{M}}{\mathrm{N}}\)
∴ \(\log _a\left(\frac{\mathrm{M}}{\mathrm{N}}\right)=x-y \text { or, } \log _a\left(\frac{\mathrm{M}}{\mathrm{N}}\right)=\log _a \mathrm{M}-\log _a \mathrm{~N}\)
∴ \(\log _a\left(\frac{M}{N}\right)=\log _a M-\log _a N\) (proved)
Corollary – 2: If n be any positive integer, then
\(\log _a\left(\frac{\mathrm{M}_1 \mathbf{M}_2 \mathrm{M}_3 \cdots \cdots \cdots \mathrm{M}_n}{\mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~N}_3 \cdots \cdots \cdots \mathrm{N}_n}\right)\)
= \(\log _a\left(\mathrm{M}_1 \mathrm{M}_2 \mathrm{M}_3 \cdots \cdots \cdots \mathrm{M}_n\right)-\log _a\left(\mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~N}_3 \cdots \cdots \cdot \mathrm{N}_n\right)\)
= \(\begin{aligned}
\left(\log _a \mathrm{M}_1+\log _a \mathrm{M}_2+\log _a \mathrm{M}_3+\cdots \cdots \cdots \cdots \cdots+\log _a \mathrm{M}_n\right)- \\
\left(\log _a \mathrm{~N}_1+\log _a \mathrm{~N}_2+\log _a \mathrm{~N}_3+\cdots \cdots \cdots \cdots \cdots+\log _a \mathrm{~N}_n\right)
\end{aligned}\)
Formula 3: \(\log _a M^n=n \log _a M\)
Proof:
Let x = \(\log _a M^n \text { and } y=\log _a M\)
∴ \(a^x=\mathrm{M}^n \text { and } a^y=\mathrm{M}\)
Now, \(a^x=\mathrm{M}^n=\left(a^y\right)^n=a^{n y}\) [ ∵ m = ay ]
∴ \(x=n y \text { or, } \log _a \mathrm{M}^n=n \log _a \mathrm{M}\)
∴ \(\log _a \mathrm{M}^n=n \log _a \mathrm{M}\) (proved)
Formula 4: \(\log _a \mathbf{M}=\log _b \mathrm{M} \times \log _a b\)
Proof:
Let x = \(\log _a M \text { and } y=\log _b M\)
∴ ax = M…………(1) and by = M…………..(2)
Now, from (1) and (2) we get by the method of comparison, \(\)
∴ \(\frac{x}{y}=\log _a b\) (by definition)
or, \(\frac{\log _a M}{\log _b M}=\log _a b\)
or, \(\log _a \mathrm{M}=\log _b \mathrm{M} \times \log _a b\)
∴ \(\log _a \mathrm{M}=\log _b \mathrm{M} \times \log _a b\) (proved)
Corollary 3: \(\log _a b \times \log _b a=\mathrm{I}\)
Proof:
Putting M = a in Formula – 4, we get, \(\log _a a=\log _b a \times \log _a b\).
or, \(1=\log _a b \times \log _b a\) [ ∵ \(\log _a b \times \log _b a=1\) ]
∴ \(\log _a b \times \log _b a=1\) (Proved).
Corollary 4: \(\log _a b=\frac{1}{\log _b a}\) [From Corollary-3]
Corollary 5: \(\log _b \mathrm{M}=\frac{\log _a \mathrm{M}}{\log _a b}\) [From formula 4]
Corollary 6: \(\log _a \mathrm{M}=\frac{\log _b \mathrm{M}}{\log _b a}\) [From formula 4 and Corollary 4]
Corollary 7: \(\frac{\log _a \mathrm{M}}{\log _a \mathrm{~N}}=\frac{\log _b \mathrm{M}}{\log _b \mathrm{~N}}\)
Proof:
By formula – 4, \(\log _a M=\log _b M \times \log _a b \text { or, } \frac{\log _a M}{\log _a b}=\log _b M\)
or,\(\frac{\log _a \mathrm{M}}{\log _a \mathrm{~N}}=\log _b \mathrm{M}\)
[ ∵ \(\log _a \mathrm{~N}=\log _b \mathrm{~N} \times \log _a b\) ]
or, \(\frac{\log _a M \times \log _b N}{\log _a N}=\log _b M \text { or, } \frac{\log _a M}{\log _a N}=\frac{\log _b M}{\log _b N}\) [Dividing by log,N]
or, \(\frac{\log _a \mathrm{M}}{\log _a \mathrm{~N}}=\frac{\log _b \mathrm{M}}{\log _b \mathrm{~N}}\) (Proved).
Corollary – 8: If \(\log _a \mathrm{M}=\log _a \mathrm{~N}, \text { then } \mathrm{M}=\mathrm{N}\).
Proof:
Let, \(\log _a \mathrm{M}=\log _a \mathrm{~N}\)
∴ \(x=\log _a \mathrm{M} \Rightarrow a^x=\mathrm{M}\)……..(1)
Again x = \(x=\log _a \mathrm{~N} \Rightarrow a^x=\mathrm{N}\)………(2)
Now, from (1) and (2) we get, M = N. (Proved)
Alternative Method:
\(\log _a \mathrm{M}=\log _a \mathrm{~N}\)∴\(a^{\log _a \mathrm{M}}, a^{\log _a \mathrm{~N}} \text { or, } \mathrm{M}=\mathrm{N}\) (proved)
[ ∵ \(\begin{aligned}
& a^{\log _a \mathrm{M}}=\mathrm{M} \\
& a^{\log _a \mathrm{~N}}=\mathrm{N}
\end{aligned}\) ]
Corollary 9: \(\log _a 1\) = 0
Proof:
We know, aº = 1,
∴ by definition of logarithm,\(\log _a 1\) = 0.
Corollary – 10: \(\log _a a=1\)
Proof:
We know, a¹ = a.
by definition of logarithm, 1 = \(\log _a a\)
∴ \(\log _a a=1\) =1 (Proved).
Corollary 11: \(a^{\log _a \mathrm{M}}=\mathrm{M}\).
Proof:
Let, x = \(x=\log _a \mathrm{M}\),
∴ ax = M …………..(1)
Now, \(a^{\log _a \mathrm{M}}=a^x=\mathrm{M}\) = M [by (1)],
∴ \(a^{\log _a \mathrm{M}}=\mathrm{M}\). (Proved).
Corollary – 12: \(\log _{a^n} \mathrm{M}=\frac{1}{n} \log _a \mathrm{M}\)
Proof:
Let, x = \(\log _{a^n} \dot{\mathbf{M}} \text { and } y=\log _a \mathbf{M}\)
∴\(\left(a^n\right)^x=\mathrm{M}\)……..(1) and ay = M………..(2)
Now, from (1) and (2) we get, \(a^{n x}=a^y \text { or, } n x=y \text { or, } x=\frac{1}{n} y \text { or, } \log _{a^n} \mathrm{M}=\frac{1}{n} \log _a \mathrm{M}\)
∴\(\log _{a^n} \dot{\mathbf{M}} \text { and } y=\log _a \mathbf{M}\) (Proved)
Corollary 13 \(\log _a\left(\frac{1}{a}\right)=-1\)
Proof :
LHS = \(\log _a\left(\frac{1}{a}\right)=\log _a(1)-\log _a a\) [by formula – 2)
= 0 – 1 [by Cor. 9 and Cor.-10 ]
= – 1
= RHS.
∴ \(\log _a\left(\frac{1}{a}\right)=-1\)
Alternative Method:
\(\log _a\left(\frac{1}{a}\right)=\log _a(a)^{-1}=-1 \cdot \log _a a\) (by formula – 3)
= – 1 x 1 (by Cor.-10)
= -1 (proved)
Corollary 14: \(\log _{a^n} \mathrm{M}^m=\frac{m}{n} \log _a \mathrm{M}\)
Proof:
Let, an = b.
∴ \(\log _{a^n} \mathrm{M}^m=\log _b \mathrm{M}^m=m \log _b \mathrm{M}\) [by formula – 3]
= \(m \log _{a^n} \mathrm{M}\) [∵ b = an]
= \(m \times \frac{1}{n} \log _a \mathrm{M}\) [by Cor.- 13]
= \(\frac{m}{n} \log _a M\)
∴\(\log _{a^n} \mathrm{M}^m=\frac{m}{n} \log _a \mathrm{M}\) (Proved)
For example: \(\log _{2 \sqrt{2}} 256=\log _{2^{\frac{3}{2}}}\left(2^8\right)\)
[ ∵ \(2 \sqrt{2}=2^1 \times 2^{\frac{1}{2}}=2^{1+\frac{1}{2}}=2^{\frac{3}{2}} \text { and } 256=2^8\) ]
= \(\frac{8}{\frac{3}{2}} \log _2 2\) [ by Cor. – 14]
= \(8 \times \frac{2}{3} \times 1=\frac{16}{3}=5 \frac{1}{3}\)
Chapter 7 Logarithm Select The Correct Answer (MCQ)
Question 1.
1. If \(\log _x \frac{1}{3}=-\frac{1}{3}\) then x =
- 27
- 9
- 3
- 1/27
Solution:
\(\log _x \frac{1}{3}=-\frac{1}{3}\)∴ \((x)^{-\frac{1}{3}}=\frac{1}{3}\)
Or, \((x)^{-\frac{1}{3} x-3}=\left(\frac{1}{3}\right)^{-3}\)
or, \(x=3^{-1 \times-3}=3^3\)
= 27
∴ 1. 27 is correct.
2. If \(\log _{\sqrt{2}} x=a, \text { then } \log _{2 \sqrt{2}} x\) =
- a/3
- 1/a
- 2a
- 3a
Solution:
\(\log _{\sqrt{2}} x=a\)∴ \((\sqrt{2})^a=x\)
∴ \(\log _{2 \sqrt{2}} x=\log _{2 \sqrt{2}}(\sqrt{2})^a\)
= \(\log _{2^{\frac{3}{2}}}\left(2^{\frac{1}{2}}\right)^a\left[ ∵ 2 \sqrt{2}=2^{\frac{3}{2}}\right]\)
= \(\log _{2^{\frac{3}{2}}}(2)^{\frac{a}{2}}=\frac{\frac{a}{2}}{\frac{3}{2}} \log _2 2 \text { [by Cor-14] }\)
= \(\frac{a}{2} \times \frac{2}{3} \times 1=\frac{a}{3}\)
∴ 1. a/3 is correct
3. If \(\log _2 3=a, \text { then } \log _8 27\)
- 3a
- 1/a
- 2a
- a
Solution:
\(\log _8 27=\log _2(3)^3\) [∵ 8 = 2³ and 27 = (3)³]
= \(\frac{3}{3} \log _2 3=\log _2 3=a\)
∴ 4. a is correct
4. If \(\log _{10}(7 x-5)=2\) =
- 10
- 12
- 15
- 18
Solution:
\(\log _{10}(7 x-5)=2\)∴ 10² = 7x – 5
or, 100 = 7x – 5
or, 7x = 105
or, x = 15.
∴ 3. 15 is correct.
5. If \(\log _{\sqrt{x}} 0 \cdot 25=4, \text { then } x\) =
- 0.5
- 0.25
- 4
- 16
Solution:
\(\log _{\sqrt{x}} 0 \cdot 25=4\)or, \((\sqrt{x})^4\)
or, x² = 0.25
or, x = √0.25
or, x = 0.5
6. \(\log _a \log _a \log _a a^{a^{a^a}}\)
- 1
- a
- aa
- cannot be determined
Solution:
\(\log _a \log _a \log _a a^{a^{a^a}}=\log _a \log _a a \log _a a^{a^a}\left[∵ \log _a a^{a^{a^a}}=a \log _a a^{a^a}\right]\)= \(\log _a \cdot 1 \cdot \log _a a^{a^a}\left[∵ \log _a a=1\right]=\log _a \log _a a^{a^a}=\log _a a \cdot \log _a a^a\)
= \(\text { 1. } \log _a a^a=\log _a a^a=a \log _a a=a .1=a\)
∴ 2. a is correct
7. \(\log _b a \times \log _c b \times \log _a c\)
- a
- b
- c
- 1
Solution:
\(\log _b a \times \log _c b \times \log _a c=\log _b a \times \log _c b \times \log _b c \times \log _a b \text { [by changing the base] }\)= \(\left(\log _b a \times \log _a b\right) \times\left(\log _c b \times \log _b c\right)\)
= 1 x1
= 1
∴ 4. 1 is correct
8. \((4)^{\log _9 3}\)
- 2
- 3
- 4
- √2
Solution:
\((4)^{\log _9 3}=(4)^{\log _{3^2} 3}=(4)^{\frac{1}{2} \log _3 3}=(4)^{\frac{1}{2} \times 1}=(4)^{\frac{1}{2}}=2\)∴ 1. 2 is correct
9. \(\log _{9 \sqrt{3}}(0 . \dot{1})\)
- 4/5
- \(-\frac{4}{5}\)
- 3/4
- \(-\frac{3}{4}\)
Solution:
\(\log _{9 \sqrt{3}}(0 . i)=\log _{9 \sqrt{3}}\left(\frac{1}{9}\right)=\log _{3^2}(3)^{-2}=\frac{-2}{\frac{5}{2}} \log _3 3=-\frac{4}{5} \times 1=-\frac{4}{5}\)∴ 2. \(-\frac{4}{5}\) is correct
10. If \(\log _x(\sqrt{5})=-\frac{1}{6}\), then x =
- 0.04
- 0.08
- 0.008
- 0.01
Solution:
\(\log _x(\sqrt{5})=-\frac{1}{6}\)∴ \((x)^{-\frac{1}{6}}=\sqrt{5} \text { or, } x^{-\frac{1}{6}}=5^{\frac{1}{2}}\)
or, \(x^{-\frac{1}{6} \times 6}=5^{\frac{1}{2} \times 6} \text { or, } x^{-1}=5^3\)
or, \(\frac{1}{x}=125 \text { or; } x=\frac{1}{125}\)
or, x = 0.008
∴ 3. 0.008 is correct
Chapter 7 Logarithm Short Answer Type Questions
Question1.
1 Find the value of \(\log _4 \log _4 \log _4 256\)
Solution:
\(\log _4 \log _4 \log _4 256=\log _4 \log _4 \log _4(4)^4\)
= \(\log _4 \log _4 4 \log _4 4=\log _4(1.1)=\log _4 1=0\)
2. Find the value of \(\log \frac{a^n}{b^n}+\log \frac{b^n}{c^n}+\log \frac{c^n}{a^n}\)
Solution:
\(\log \frac{a^n}{b^n}+\log \frac{b^n}{c^n}+\log \frac{c^n}{a^n}\)
= \(\log a^n-\log b^n+\log b^n-\log c^n+\log c^n-\log a^n=0\)
3. Show that \(a^{\log _a x}=x\).
Solution:
Let \(a^{\log _a x}=p \text { and } \log _a x=q\)
∴ aq = p .……….(1) and aq = x……. (2)
From (1) and (2) we get, p = x
∴ \(a^{\log _a x}=x\)(Proved).
4. If \(\log _e 2 \cdot \log _x 25=\log _{10} 16 \cdot \log _e 10\), then find the value of x.
Solution:
\(\log _e 2 \cdot \log _x 25=\log _{10} 16 \cdot \log _e 10 \text { or, } \log _e 2 \cdot \log _x 5^2=\log _{10} 2^4 \log _e 10\)
or,\(\log _e 2.2 \log _x 5=4 \log _{10} 2 \cdot \log _e 10 \text { or, } \log _e 2 \cdot \log _x 5=2 \log _{10} 2 \cdot \log _e 10\)
or, \(\log _e 2 \cdot \log _x 5=2 \log _e 2 \cdot \log _{10} e \cdot \log _e 10\)
or, \(\log _x 5=2 \text { or, } x^2=5 \text { or, } x=\sqrt{5}\)
∴ the required value of x = √5.
5. Show that \(\log _b\left(\frac{1}{b^n}\right)=-n\)
Solution:
LHS = \(\log _b\left(\frac{1}{b^n}\right)=\log _b 1-\log _b b^n\) [by formula – 2]
= \(\left.0-n \log _b b \text { [by formula }-3\right]\) = -n [∵ \(\log _b b=1\) ] = -n
= RHS (Proved).
6. If \(\log _{10} x-\log _{10} \sqrt{x}=\frac{2}{\log _{10} x}\) then find the value of x.
Solution:
\(\log _{10} x-\log _{10} \sqrt{x}=\frac{2}{\log _{10} x} \text { or, } \log _{10} x-\frac{1}{2} \log _{10} x=\frac{2}{\log _{10} x}\)
[∵ \(\sqrt{x}=x^{\frac{1}{2}}\) ]
or, \(\frac{1}{2} \log _{10} x=\frac{2}{\log _{10} x}\)
or, \(\left(\log _{10} x\right)^2=4 \text { or, } \log _{10} x= \pm 2\)
∴ \(\log _{10} x=2 \text { or, } 10^2=x \text { or, } x=100\)
Again, \(\log _{10} x=-2 \text { or, } 10^{-2}=x \text { or, } x=\frac{1}{100}\)
∴ \(x=100 \text { or } \frac{1}{100}\)
7. Show that \(\log _3 \log _2 \log _{\sqrt{3}} 81=1\)
Solution:
LHS = \(\log _3 \log _2 \log _{\sqrt{3}} 81=\log _3 \log _2 \log _{\frac{1}{3}}(3)^4\)
= \(=\log _3 \log _2\left(\frac{4}{\frac{1}{2}}\right) \log _3 3=\log _3 \log _2 8.1 \quad\left[ ∵ \frac{4}{\frac{1}{2}}=8 \text { and } \log _3 3=1\right]\)
= \(\log _3 \log _2 2^3=\log _3 .3 \log _2 2=\log _3 3.1=\log _3 3\)
= 1
= RHS (proved)
8. Show that \(\log _b a \times \log _c b \times \log _d c=\log _d a\)
Solution:
LHS = \(\log _b a \times \log _c b \times \log _d c=\left(\log _c b \times \log _b a\right) \times \log _d c\)
= \(\log _c a \times \log _d c=\log _d c \times \log _c a=\log _d a\)
= RHS (proved)
9. Find the value of \((y z)^{\log y-\log z} \times(z x)^{\log z-\log x} \times(x y)^{\log x-\log y}\)
Solution:
\((y z)^{\log y-\log z} \times(z x)^{\log z-\log x} \times(x y)^{\log x-\log y}\)= \(x^{\log z-\log x+\log x-\log y} \cdot y^{\log y-\log z+\log x-\log y} \cdot z^{\log y-\log z+\log z-\log x}\)
= \(=x^{\log z-\log y} \cdot y^{\log x-\log z} \cdot z^{\log y-\log x}=x^{\log \frac{z}{y}} \cdot y^{\log \frac{x}{z}} \cdot z^{\log \frac{y}{x}}\)………….(1)
Now let, u = \(x^{\log \frac{z}{y}} \cdot y^{\log \frac{x}{z}} \cdot z^{\log \frac{y}{x}}\)
∴ \(=\log \left(x^{\log \frac{z}{y}} \cdot y^{\log \frac{x}{z}} \cdot z^{\log \frac{y}{x}}\right)\)
or, \(\log u=\log x^{\log \frac{z}{y}}+\log y^{\log \frac{x}{z}}+\log z^{\log \frac{y}{x}}\)
or, \(\log u=\log \frac{z}{y}+\log \frac{x}{z}+\log \frac{y}{x}\)
or, \(\log u=\log z-\log y+\log x-\log z+\log y-\log x\)
or, \(\log u=0 \Rightarrow u=1\)
∴ the value of the given expression = 1.
10. Prove that \(a^{\log _{a^2} x} \cdot b^{\log _{b^2}{ }^y} \cdot c^{\log _{c^2} z}=\sqrt{x y z}\)
Solution:
LHS = \(a^{\log _a 2 x} \cdot b^{\log _{b^2} y} \cdot c^{\log _{c^2} z}=a^{\frac{1}{2} \log _a x} \cdot b^{\frac{1}{2} \log _b y} \cdot c^{\frac{1}{2} \log _c z}\)
= \(a^{\log _a x^{\frac{1}{2}}} \cdot b^{\log _b y^{\frac{1}{2}}} \cdot c^{\log _c z^{\frac{1}{2}}}=x^{\frac{1}{2}} \cdot y^{\frac{1}{2}} \cdot z^{\frac{1}{2}}=\sqrt{x y z}\)
=RHS
Chapter 7 Logarithm Long Answer Type Questions
Question 1. If x, y, and z be three consecutive integers, then prove that log (1 + xz) = 2 log y.
Solution:
Given x, y, and z Be Three Consecutive Integers:-
Let x, y, and z be three consecutive integers.
y=x+1 and z = y + 1 = x + 1 + 1 = x+2.
Now,
LHS = log (1 + xz) = log [1 + x (x + 2)] [∵ z = x + 2]
= log (1 + x² + 2x)
= log (x + 1)²
=2 log (x + 1).
= 2 log y [∵ x+1 = y] = RHS (Proved).
Question 2.
1. If \(1+\log _{10} a=2 \log _{10} b\), then express a in terms of b².
Solution:
\(1+\log _{10} a=2 \log _{10} b \text { or, } \log _{10} 10+\log _{10} a=\log _{10} b^2\)or, \(\log _{10}(10 a)=\log _{10} b^2\)
∴ \(10 a=b^2 \text { or, } a=\frac{b^2}{10}\)
2. If \(3+\log _{10} x=2 \log _{10} y\), then express x in terms of y.
Solution:
\(3+\log _{10} x=2 \log _{10} y \text { or, } 3 \log _{10} 10+\log _{10} x=\log _{10} y^2\)or, \(\log _{10} 10^3+\log _{10} x=\log _{10} y^2 \text { or, } \log _{10}(1000 x)=\log _{10} y^2\)
⇒ \(1000 x=y^2 \text { or, } x=\frac{y^2}{1000}\)
Question 3.
1. \(\frac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2}\)
Solution:
\(\frac{\log \sqrt{27}+\log 8-\log \sqrt{1000}}{\log 1.2}=\frac{\log \left(\frac{\sqrt{27} \times \sqrt{64}}{\sqrt{1000}}\right)}{\log 1.2}[ ∵ 8=\sqrt{64}]\)= \(\frac{\log \left(\frac{\sqrt{3^3} \times \sqrt{4^3}}{\sqrt{10^3}}\right)}{\log 1.2}=\frac{\log \left(\frac{3 \times 4}{10}\right)^{\frac{3}{2}}}{\log 1.2}\)
= \(\frac{\frac{3}{2} \log \frac{12}{10}}{\log \frac{12}{10}}=\frac{3}{2}\)
2. \(\log _3 4 \times \log _4 5 \times \log _5 6 \times \log _6 7 \times \log _7 3\)
Solution:
\(\log _3 4 \times \log _4 5 \times \log _5 6 \times \log _6 7 \times \log _7 3\)= \(\log _3 5 \times \log _5 6 \times \log _6 7 \times \log _7 3\)
= \(\log _3 6 \times \log _6 7 \times \log _7 3=\log _3 7 \times \log _7 3=1\)
3. \(\log _{10}\left(\frac{384}{5}\right)+\log _{10}\left(\frac{81}{32}\right)+3 \log _{10}\left(\frac{5}{3}\right)+\log _{10}\left(\frac{1}{9}\right)\)
Solution:
\(\log _{10}\left(\frac{384}{5}\right)+\log _{10}\left(\frac{81}{32}\right)+3 \log _{10}\left(\frac{5}{3}\right)+\log _{10}\left(\frac{1}{9}\right)\)= \(\log _{10}\left(\frac{384}{5} \times \frac{81}{32} \times \frac{5^3}{3^3} \times \frac{1}{9}\right)\)
=\(\log _{10}(100)=\log _{10}(10)^2=2 \log _{10} 10=2 \times 1=2\)
4. \(\log _{x^2} x \times \log _{y^2} y \times \log _{z^2} z\)
Solution:
\(\log _{x^2} x \times \log _{y^2} y \times \log _{z^2} z\)= \(\frac{1}{2} \log _x x \times \frac{1}{2} \log _y y+\frac{1}{2} \log _z z \quad\left[∵ \log _{a^n} M=\frac{1}{n} \log _a \mathrm{M}\right]\)
= \(=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8} \quad\left[∵ \log _x x=\log _y y=\log _z z=1\right]\)
Question 6. Prove that
1. \(\log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}=\log 2\)
Solution:
= \(\log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}=\log \left(\frac{75}{16} \div \frac{5^2}{9^2} \times \frac{32}{243}\right)\)
= \(\log \left(\frac{75}{16} \times \frac{81}{25} \times \frac{32}{243}\right)\)
= log2
= RHS (proved).
2. \(\log _{10} 15\left(1+\log _{15} 30\right)+\frac{1}{2} \log _{10} 16\left(1+\log _4 7\right)-\log _{10} 6\left(\log _6 3+1+\log _6 7\right)=2\)
Solution:
= \(\log _{10} 15\left(1+\log _{15} 30\right)+\frac{1}{2} \log _{10} 16\left(1+\log _4 7\right)-\log _{10} 6\left(\log _6 3+1+\log _6 7\right)\)
= \(\begin{aligned}
&\log _{10} 15+\log _{10} 15 \times \log _{15} 30+\frac{1}{2} \log _{10} 4^2\left(1+\log _4 7\right) \\
& -\log _{10} 6 \times \log _6 3-\log _{10} 6-\log _{10} 6 \times \log _6 7
\end{aligned}\)
= \(\log _{10} 15+\log _{10} 30+\frac{1}{2} \times 2 \log _{10} 4+\frac{1}{2} \times 2 \log _{10} 4 \times \log _4 7-\log _{10} 3-\log _{10} 6-\log _{10} 7\)
= \(\log _{10} 15+\log _{10} 30+\log _{10} 4+\log _{10} 7-\log _{10} 3-\log _{10} 6-\log _{10} 7\)
= \(\log _{10} 15+\log _{10} 30+\log _{10} 4-\log _{10} 3-\log _{10} 6\)
= \(\log _{10}\left(\frac{15 \times 30 \times 4}{3 \times 6}\right)=\log _{10} 100=\log _{10}(10)^2\)
= \(2 \log _{10} 10\)
= 2.1
= 2
= RHS (proved)
3. \(\log _2 \log _2 \log _4 256+2 \log _{\sqrt{2}} 2=5\)
Solution:
= \(\log _2 \log _2 \log _4 256+2 \log _{\sqrt{2}} 2=\log _2 \log _2 \log _4(4)^4+2 \log _{2^{\frac{1}{2}}}(2)\)
= \(\log _2 \log _2 4 \log _4 4+2 \times \frac{1}{\frac{1}{2}} \log _2 2=\log _2 \log _2(2)^2 \cdot 1+2 \times 2 \times 1\)
= \(\log _2 2 \log _2 2+4=\log _2 2 \times 1+4\)
= 1 + 4
= 5
= RHS (proved)
4. \(\log _b 3 a \times \log _c 3 b \times \log _a 3 c=\frac{1}{27}\)
Solution:
= \(\log _{b^3} a \times \log _c 3 b \times \log _{a^3} c=\frac{1}{3} \log _b a \times \frac{1}{3} \log _c b \times \frac{1}{3} \log _a c\)
= \(\frac{1}{27}\left(\log _c b \times \log _b a\right) \times \log _a c\)
= \(\frac{1}{27} \log _c a \times \log _a c=\frac{1}{27} \times 1=\frac{1}{27}\)
= RHS (proved)
5. \(\frac{1}{\log _{x y}(x y z)}+\frac{1}{\log _{y z}(x y z)}+\frac{1}{\log _{z x}(x y z)}=2\)
Solution:
= \(\frac{1}{\log _{x y}(x y z)}+\frac{1}{\log _{y z}(x y z)}+\frac{1}{\log _{z x}(x y z)}\)
= \(\log _{x y z}(x y)+\log _{x y z}(y z)+\log _{x y z}(z x)\)
= \(\log _{x y z}(x y \times y z \times z x)\)
= \(\log _{x y z}(x y z)^2\)
= \(2 \log _{x y z}(x y z)\)
= 2 x 1
= 2
= RHS (proved)
6. \(\log \frac{a^2}{b c}+\log \frac{b^2}{c a}+\log \frac{c^2}{a b}=0\)
Solution:
= \(=\log \frac{a^2}{b c}+\log \frac{b^2}{c a}+\log \frac{c^2}{a b}\)
= \(\log \left(\frac{a^2}{b c} \times \frac{b^2}{c a} \times \frac{c^2}{a b}\right)\)
= \(\log \left(\frac{a^2 b^2 c^2}{a^2 b^2 c^2}\right)\)
= log 1
= 0
= RHS (proved)
7. \(x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=1\)
Solution:
Let,
u = \(x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}\)
⇒ log u = \(\log \left[x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}\right]\)
= \(\log x^{\log y-\log z}+\log y^{\log z-\log x}+\log z^{\log x-\log y}\)
= \((\log y-\log z) \log x+(\log z-\log x) \log y+(\log x-\log y) \log z\)
∴ log u = 0
or, log u = log 1
⇒ u = 1
∴ \(x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}\) = 1. (proved)
8. \(\log _7 \sqrt{7 \sqrt{7 \sqrt{7 \ldots \ldots \ldots \infty \infty}}}=1\)
Solution:
Let, \(\sqrt{7 \sqrt{7 \sqrt{7 \cdots \cdots \cdots \cdots \infty}}}=x\)
or, \(7 \sqrt{7 \sqrt{7 \sqrt{7 \cdots \cdots \cdots \cdots \infty}}}\) = x²
or, 7x = x² [∵ \(\sqrt{7 \sqrt{7 \sqrt{7 \cdots \cdots \cdots \cdots \infty}}}=x\) ]
or, x² – 7x = 0
or, x(x – 7) = 0
∴ either x = 0
or, x – 7 = 0
or, x = 7
∵ the logarithm of 0 is undefined,
∴ x ≠ 0 and x = 7
∴ \(\log _7 \sqrt{7 \sqrt{7 \sqrt{7 \cdots \cdots \cdots \cdots \infty}}}=\log _7 7\)
= 1
= RHS (proved)
Question 7.
1. If \(\log \left(\frac{x+y}{5}\right)=\frac{1}{2}(\log x+\log y), \text { then show that } \frac{x}{y}+\frac{y}{x}=23\)
Solution:
Given that \(\log \left(\frac{x+y}{5}\right)=\frac{1}{2}(\log x+\log y)\)
or, \(2 \log \left(\frac{x+y}{5}\right)=\log x+\log y\)
or, \(\log \left(\frac{x+y}{5}\right)^2=\log (x y)\)
∴ \(\left(\frac{x+y}{5}\right)^2=x y\)
or, \((x+y)^2=25 x y \text { or, } x^2+2 x y+y^2=25 x y \text { or, } x^2+y^2=23 x y\)
or, \(\frac{x^2}{x y}+\frac{y^2}{x y}=23\) [Dividing by xy]
or, \(\frac{x}{y}+\frac{y}{x}=23\)
∴ \(\frac{x}{y}+\frac{y}{x}=23\) (proved)
2. If \(a^4+b^4=14 a^2 b^2\) , then show that \(\left(a^2+b^2\right)=\log a+\log b+2 \log 2\)
Solution:
Given that \(a^4+b^4=14 a^2 b^2, \text { then }\left(a^2+b^2\right)^2-2 a^2 b^2=14 a^2 b^2]\)
or, \(\left(a^2+b^2\right)^2=16 a^2 b^2\)
\(\left(a^2+b^2\right)^2=(4 a b)^2 \text { or, } a^2+b^2=4 a b\) [ by taking square root]
∴ \(\log \left(a^2+b^2\right)=\log (4 a b)=\log 4+\log a+\log b\)
= \(\log 2^2+\log a+\log b=2 \log 2+\log a+\log b\)
∴ \(\log \left(a^2+b^2\right)=\log a+\log b+2 \log 2\) (proved)
Question 8. If \(\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}\), then prove that xyz = 1.
Solution:
Let, \(\frac{\log x}{y-z}=\frac{\log y}{z-x}=\frac{\log z}{x-y}\) = k
∴ \(\frac{\log x}{y-z}=k \text { or, } \log x=k(y-z)\)……………(1)
\(\frac{\log y}{z-x}=k \text { or, } \log y=k(z-x)\)…………(2)
\(\frac{\log z}{x-y}=k \text { or, } \log z=k(x-y)\)……………(3)
Now, adding (1),(2) we get, log x + log y + lo x = k(y-z) + k(z-x) + k(x-y)
or, log(xyz) = k(y-z+z-x+z-y) = k x 0
or, log(xyz) = 0
or, log(xyz) = log 1 [∵ log 1 = 0]
∴ xyz = 1 (proved)
Question 9. If \(\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}\), then proved that
1. \(x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=1\)
2. \(x^{b^2+b c+c^2} \cdot y^{c^2+c a+a^2} \cdot z^{a^2+a b+b^2}=1\)
Solution:
Let, \(\frac{\log x}{b-c}=k \quad \text { or, } \log x=k(b-c)\)
∴ \(\frac{\log x}{b-c}=k \quad \text { or, } \log x=k(b-c)\)……………(1)
\(\frac{\log y}{c-a}=k \text { or, } \log y=k(c-a)\)……………(2)
\(\frac{\log z}{a-b}=k \text { or, } \log z=k(a-b)\)……………(3)
1. \(x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=1\)
Solution:
\(\log \left[x^{b+c} \cdot y^{c+a} \cdot z^{a+b}\right]\)= \(\log x^{b+c}+\log y^{c+a}+\log z^{a+b}\) [∵ log (MNP) = logM + logN + logP]
= \((b+c) \log x+(c+a) \log y+(a+b) \log z\)
= \((b+c) \cdot k(b-c)+(c+a) k(c-a)+(a+b) k(a-b)[\text { from }(1),(2), \text { and (3)] }\)
= \(k\left(b^2-c^2\right)+k\left(c^2-a^2\right)+k\left(a^2-b^2\right)=k\left(b^2-c^2+c^2-a^2+a^2-b^2\right)\)
= k x 0
= 0
log 1 [∵ log 1 = 0]
∴ \(\log \left[x^{b+c} \cdot y^{c+a} \cdot z^{a+b}\right]=\log 1\)
∴ \(x^{b+c} \cdot y^{c+a} \cdot z^{a+b}=1\) (proved)
2. \(x^{b^2+b c+c^2} \cdot y^{c^2+c a+a^2} \cdot z^{a^2+a b+b^2}=1\)
Solution:
\(\log \left[x^{b^2+b c+c^2} \cdot y^{c^2+c a+a^2} \cdot z^{a^2+a b+b^2}\right]\)= \(\log x^{b^2+b c+c^2}+\log y^{c^2+c a+a^2}+\log z^{a^2+a b+b^2}\)
= \(\left(b^2+b c+c^2\right) \log x+\left(c^2+c a+a^2\right) \log y+\left(a^2+a b+b^2\right) \log z\)
= \(\left(b^2+b c+c^2\right) k(b-c)+\left(c^2+c a+a^2\right) k(c-a)+\left(a^2+a b+b^2\right)\)
k(a-b)[\text { from (1), (2) and (3)]
= \(k\left(b^3-c^3\right)+k\left(c^3-a^3\right)+k\left(a^3-b^3\right)\)
= \(k\left(b^3-c^3+c^3-a^3+a^3-b^3\right)=k \times 0=0=\log 1\) [∵ log1 = 0]
∴ \(\log \left[x^{b^2+b c+c^2} \cdot y^{c^2+c a+a^2} \cdot z^{a^2+a b+b^2}\right]=\log 1\)
∴ \(x^{b^2+b c+c^2} \cdot y^{c^2+c a+a^2}, z^{a^2+a b+b^2}=1\) (proved)
Question 10. If \(a^{3-x} \cdot b^{5 x}=a^{5+x} \cdot b^{3 x} \text {, then show that } x \log \left(\frac{b}{a}\right)=\log a\)
Solution:
Given that \(a^{3-x} \cdot b^{5 x}=a^{5+x} \cdot b^{3 x} \text { or, } \frac{a^{3-x}}{a^{5+x}}=\frac{b^{3 x}}{b^{5 x}}\)
or, \(a^{3-x-5-x}=b^{3 x-5 x} \text { or, } a^{-2 x-2}=b^{-2 x} \text { or, } a^{-2(x+1)}=b^{-2 x} \text { or, } a^{x+1}=b^x\)
∴ \(\log a^{x+1}=\log b^x \text { or, }(x+1) \log a=x \log b \text { or, } x \log a+\log a=x \log b\)
or, \(\log a=x \log b-x \log a \text { or, } \log a=x(\log b-\log a) \text { or, } \log a=x \log \left(\frac{b}{a}\right)\)
∴ \(x \log \left(\frac{b}{a}\right)=\log a\) (proved)
Question 11. Show that the value of \(\log _{10} 2\) lies in between \(\frac{1}{4}\) and \(\frac{1}{3}\).
Solution:
We know, \(8<10<16 \text { or, } 2^3<10<2^4\)
∴ \(\log _2 2^3<\log _2 10<\log _2 2^4\)
or, \(3 \log _2 2<\log _2 10<4 \log _2 2 \text { or, } 3<\log _2 10<4\) [∵ \(\log _2 2=1\) ]
or, \(\frac{1}{3}>\frac{1}{\log _2 10}>\frac{1}{4} \text { or, } \frac{1}{3}>\log _{10} 2>\frac{1}{4}\) [∵ \(\frac{1}{\log _2 10}=\log _{10} 2\) ]
∴ \(\frac{1}{4}<\log _{10} 2<\frac{1}{3}\)
∴ the value of \(\) lies in between \(\log _{10} 2\) lies in between \(\frac{1}{4}\) and \(\frac{1}{3}\) (proved).
Question 12.
1. \(\log _8\left[\log _2\left\{\log _3\left(4^x+17\right)\right\}\right]=\frac{1}{3}\)
Solution:
\(\log _8\left[\log _2\left\{\log _3\left(4^x+17\right)\right\}\right]=\frac{1}{3}\) = \(\log _2 3\left[\log _2\left\{\log _3\left(4^x+17\right)\right\}\right]=\frac{1}{3}\)
or, \(\frac{1}{3} \log _2\left[\log _2\left\{\log _3\left(4^x+17\right)\right\}\right]=\frac{1}{3}\)
or, \(\log _2\left[\log _2\left\{\log _3\left(4^x+17\right)\right\}\right]=1\) [Multiplying both the sides by 3]
or, \(\log _2\left\{\log _3\left(4^x+17\right)\right\}=2^1\) [by definitiion]
or, \(\log _3\left(4^x+17\right)=(2)^{2^1}\) [by definition]
or, \(4^x+17=3^4\) [∵ \((2)^{2^1}=4\) ]
or, \(4^x+17=81 \text { or, } 4^x=64 \text { or, } 4^x=4^3\)
∴ x = 3
∴ the required solution: x = 3.
2. \(\log _8 x+\log _4 x+\log _2 x=11\)
Solution:
\(\log _8 x+\log _4 x+\log _2 x=11 \text { or, } \log _{2^3} x+\log _{2^2} x+\log _2 x=11\)or, \(\frac{1}{3} \log _2 x+\frac{1}{2} \log _2 x+\log _2 x=11\)
or, \(\log _2 x\left(\frac{1}{3}+\frac{1}{2}+1\right)=11 \text { or, } \log _2 x \times \frac{11}{6}=11\)
or, \(\log _2 x=6 \text { or, } x=2^6 \text { or, } x=64\)
∴ the required solution is: x = 64.
3. \(4^{\log _9 3}+9^{\log _2 4}=10^{\log _x 83}\)
Solution:
\(4^{\log _9 3}+9^{\log _2 4}=10^{\log _x 83}\)or, \(4^{\log _{3^2} 3}+9^{\log _2 2^2}=10^{\log _x 83} \text { or, } 4^{\frac{1}{\log _3 3}}+9^{2 \log _2 2}=10^{\log _x 83}\)
or, \(4^{\frac{1}{2}}+9^2=10^{\log _x 83}\) [∵ \(\log _3 3=\log _2 2=1\) ]
or, \(2+81=10^{\log _x 83}\)
or, \(83=10^{\log _x 83}\) ……..(1)
∴ \(\log _x 83=\log _{10} 83\)
∴ x = 10 [let \(\log _{10} 83=u, ∴ 83=10^u \cdot \text { by }\) ]
⇒ \(u=\log _{10} 83 \text { (by definition) }=\log _x 83=\log _{10} 83\)
∴ the required solutions are: x = 10.
4. \(\log _{10} x-\log _{10} \sqrt{x}=\frac{2}{\log _{10} x}\)
Solution:
\(\log _{10} x-\log _{10} \sqrt{x}=\frac{2}{\log _{10} x} \text { or, } \log _{10}\left(\frac{x}{\sqrt{x}}\right)=\frac{2}{\log _{10} x} \text { or, } \log _{10} \sqrt{x}=\frac{2}{\log _{10} x}\)or, \(\frac{1}{2} \log _{10} x=\frac{2}{\log _{10} x} \text { or, }\left(\log _{10} x\right)^2=4 \text { or, } \log _{10} x= \pm 2\)
∴ \(x=10^2 \text { or, } x=100 ; \text { Again, } x=10^{-2}=\frac{1}{100}\)
∴ the required solutions are: x = \(100, \frac{1}{100}\)
