Algebra Chapter 6 Laws Of Indices
Chapter 6 Laws Of Indices What is Index
An index is a number of a quantity by which a repeating multiplication of any number or, quantity is expressed in brief and which is written in the slightly right side on the head of the number or quantity mentioned in the first. Such as- if 2 is multiplied 4 times we write 24, i.e., 2 x 2 x 2 x 2 = 24.
Similarly, 3 × 3 × 3 = 3³, 6 x 6 x 6 x ……. (10 times) = 610, etc.
Mathematically, if a be any real number and n be any positive integer, then
an = a x a x a x ………(n-times).
Read and Learn More WBBSE Solutions For Class 9 Maths
- an is not called the index. It is a representation in the form of an index.
- n is called the index.
- a is called the base.
- Instead of a positive integer, if n is a zero, then an = 1 when a 0.
- If n be a negative integer, then a not = a x a x a x a x ……(n-times).
- Then \(a^n=a^{-m}(\text { where } n=-m, m>0)=\frac{1}{a^m}\)
= \(\frac{1}{a \times a \times a \times \ldots \ldots(m-\text { times })}\) - If n is any positive rational number, let where n = \(\frac{p}{q}\), both p and q are positive integers, and q ≠ 0.
∴ \(a^n=a^{\frac{p}{q}}=\sqrt[q]{a \times a \times a \times \ldots \ldots \ldots(p-\text { times })}\) - If n is any negative rational number, let n = – \(\frac{p}{q}\) where both p and q are positive integers and p ≠ 0.\(a^n=a^{-\frac{p}{q}}=\frac{1}{\sqrt{a^p}}=\frac{1}{\sqrt[2]{a \times a \times a \times 1 \ldots. .(p-\text { times })}}\)
- If n is any irrational number, let n√p, where p > 0 and p is not a perfect square. \(a^n=a^{\sqrt{p}}\)
Chapter 6 Laws Of Indices What Is Base And Index Or Power
By definition, an = a x a x a x……….. (n-times).
Here, a is multiplied by ‘a’ itself n times.
This ‘a’ is called the base.
Therefore, the number or quantity which is in brief represented in the form of an index instead of producing the number by itself repeatedly is called the base of the form.
For example, in index 24, the base is 2, in the index yn, the base is y,……etc.
Again in the quantity an = a x a x a x…….(n-times), a is repeatedly multiplied by itself for n-times.
This is said to be the index or power of the base.
Therefore, the number of quantities by which, how many times any base is repeatedly multiplied by itself, is expressed, is called the index or power of the base.
For example in the quantity a4, the index of a is 4, in the quantity b6, the index of b is 6, in the quantity c10, the index of c is 10, in the quantity, the index of x is y, in the quantity y-3, the index of y is (-3), in the quantity z-7, the index of z is (-7), in the quantity (y)(-z), the index of y is (-z), in the quantity (x)(-y), the index of
(-x) is (-y)……. etc.
Chapter 6 Laws Of Indices What Is Root
If a and x be any real numbers and
feature in this Lose
be any positive integer then let an = x,
∴ \(a=\sqrt[n]{x}=x^{\frac{1}{n}}\)
∴ a is called anth root of x. Since x is a real number and a=xn, a can have n-values, among which only one should be positive, i.e. if x is a real number then the number of nth roots of n will be n in number.
If n = 1, then a¹ = x
or, a = x.
If n = 2, then a² = x
or, a = √x
or, a = x1/2
If n = 3, then a3 = x
or, a=3√x
or, a=x1/3
Here, 3√x is called the cube root of x.
You have already learned that d² means a is multiplied by a for 2 times, then what is the meaning of a1/2? The answer is the meaning of a1/2 is a is multiplied by itself for ½ times. In fact, is there any real meaning to it.
Apparently, it seems that a cannot be multiplied by a for ½ times, it thus seems to be meaningless.
But mathematically it has a great significance.
The mathematical significance of a is to split an into two such equal parts the product of which is equal to a, ie, the meaning of a is to determine those values, the product of each of which by itself is always equal to a.
For example, 41/2 = (+2), since (+ 2) x (+ 2) = 4.
Again, 4 (-2), since (-2) × (-2) = 4.
∴ We can say that 4² is not meaningless and in fact 41/2 = ±2.
Similarly, the meaning of a1/3 is to determine three such values (real or imaginary), the repeated product of each of which by themselves for 3 times is always equal to a. It is very much possible mathematically.
Again, according to the representation of numbers or quantities in the form of index, the meaning of a1/3 is multiplied by itself for (-⅓) times, which may appear to us as meaningless but it is also not meaningless mathematically.
For example, \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\) which is a real value of 27-1/3.
It is possible to determine two other values of 27-1/3, but they are imaginary.
In modern mathematics real number systems and complex number systems, both are of equal significance.
Where the real number system becomes still. met, the complex number system starts from there.
From the above discussion, we get,
if 25 = 2 x 2 x 2 x 2 x 2 = 32, then 5√32 = 2,
if (-2) = -2x-2x- -2x-2x-2= 32, then 5√-32 = -2,
if 3³ = 3 x 3 x 3 = 27, then 3√27 = 3,
if an= a x a x a x…………..(n-times) = x, then n√x=a.
Chapter 6 Laws Of Indices Laws Of Indices
If a and b be any two real numbers and m and n be any two positive integers, then
Formula 1: \(a^m \times a^n=a^{m+n}\)
Formula 2: \(a^m \div a^n=\left\{\begin{array}{l}
a^{m-n}, \text { when } m>n \\
\frac{1}{a^{n-m}}, \text { when } \cdot n>m
\end{array}\right.\)
Formula 3: \(\left(a^m\right)^n=a^{m n}\)
Formula 4: \((a b)^m=a^m \cdot b^m\)
Formula 5: \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}, b \neq 0\)
Proof of the formulae:
Formula 1: \(a^m \times a^n=a^{m+n}\)
Proof:
By the definition we get, am = a x a x a x …..…(m-times)
an = a x a x a x…………. (n-times)
and \(a^{m+n}=a \times a \times a \times\)……….(m + n-times)
Now,
\(a^m \times a^n=(a \times a \times a \times \ldots \ldots \ldots \ldots \text {-times }) \times(a \times a \times a \times \ldots \ldots \ldots n \text {-times })\)= \(a \times a \times a \times \ldots \ldots \ldots(m+n) \text { times }=a^{m+n}\)
∴ \(a^m \times a^n \doteq a^{m+n}\) proved
Formula 2: \(a^m \div a^n=\left\{\begin{array}{l}
a^{m-n}, \text { when } m>n \\
\frac{1}{a^{n-m}}, \text { when } \cdot n>m
\end{array}\right.\)
Proof:
By the definition we get,
\(a^m=a \times a \times a \times\)…………(m times)
\(a^n=a \times a \times a \times\)…………(n times)
\(a^{m-n}=a \times a \times a \times……………..(m-n)\) (m – n) times
Now,
\(a^m \div a^n=\frac{a^m}{a^n}=\frac{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots .(m \text { times })}{a \times a \times a \times \ldots \ldots \ldots \ldots .(n \text { times })}\)If \(m>n \text {, then } \frac{a^m}{a^n}=\frac{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots . .(m \text { times })}{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots \ldots . .(n \text { times })}\)
\(=a \times a \times a \times(m-n) times =a^{m-n}\)If \(n>m, \text { then } \frac{a^m}{a^n}=\frac{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots(m \text { times })}{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots(n \text { times })}\)
\(\frac{1}{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots \ldots .(n-m) \text { times }}=\frac{1}{a^{n \_m}}\)∴ \(a^m \div a^n=\left\{\begin{array}{l}
a^{m-n}, \text { when } m>n \\
\frac{1}{a^{n-m}}, \text { when } n>m
\end{array}\right.\) proved
Formula 3: \(\left(a^m\right)^n=a^{m n}\)
Proof:
By definition we get,
\(\left(a^m\right)^n=a^m \times a^m \times a^m \times………………(n times)\)= \(\begin{array}{r}(a \times a \times a \times \ldots \ldots \ldots . m \text { times }) \times(a \times a \times a \times \ldots \ldots \ldots \ldots . m \text { times }) \\\quad \times(a \times a \times a \times \ldots \ldots \ldots . m \text { times }) \times \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(n \text { times })
\end{array}\)
= \(a \times a \times a \times………..(m+m+m+n times)\)
= \(a \times a \times a \times \ldots \ldots \ldots \ldots \ldots(m n) \text { times }=a^{m n}\)
∴ \(\left(a^m\right)^n=a^{m n}\) Proved
Formula 4: \((a b)^m=a^m \cdot b^m\)
Proof:
By the definition we get,
\((a b)^m=a b \times a b \times a b \times\)………….(m – times)
\(\{a \times a \times a \times \ldots \ldots \ldots . .(m \text {-times }\} \times\{b \times b \times b \times \ldots \ldots \ldots . .(m \text {-times })\}\) \(\dot{a}^m \times b^m \text { (by definition) }\)∴ \((a b)^m=a^m \cdot b^m\) proved
Formula 5: \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}, b \neq 0\)
Proof:
By definition we get,
\(\left(\frac{a}{b}\right)^{\prime \prime}=\frac{a}{b} \times \frac{a}{b} \times \frac{a}{b} \times . .\) \(\frac{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots .(m \text { times })}{b \times b \times b \times \ldots \ldots \ldots \ldots \ldots .(m \text { times })}=\frac{a^m}{b^m}\)∴\(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}, b \neq 0\) proved
Formula 1: of the laws of indices is called the Fundamental law of Indices, i,e. the formula \(a^{m \prime} \times a^n=a^{m+n}\) is the fundamental law of indices. Because all other formulae can be proved with the help of this formula.
In the following, proofs of all other formulae with the help of formula-1 are shown:
Formula 2: \(a^m \div a^n=\left\{\begin{array}{l}
a^{m-n}, \text { when } m>n \\
\frac{1}{a^{n-m}}, \text { when } \cdot n>m
\end{array}\right.\)[/latex]
Proof:
By definition, we get, m>n, m-n>0.
∴ \(a^{m-n} \times a^n=a^{m-n+n}\left[\text { by formula-1] }=a^m\right.\)
\(a^{m-n}=\frac{a^m}{a^n}=a^m \div a^n\)\(a^m \div a^n=a^{m-n}, \text { when } m>n\) proved
when n>m, n-m>0 or, m-n<0, let m-n = -p, where p is positive integer.
Now, from formula-1 we get, \(a^m \times a^n=a^{m+n} \text { or, } a^{-n} \times a^n=a^{\circ} \text { or, } a^{-n} \times a^n=1\)
∴ \(a^{-n}=\frac{1}{a^n}\)
∴ \(a^{-p}=\frac{1}{a^p}\)
Then \(a^{m-n} \times a^n=a^{m-n+n}\) [by formula 1]
or, \(a^{-p} \times a^n=a^m\)
or, \(a^{-p}=\frac{a^m}{a^n}\)
or, \(\frac{1}{a^{n-m}}=a^m \div a^n\)
∴ \(a^m \div a^n=\frac{1}{a^{n-m}}\), when n>m (proved)
Formula 3: \(\left(a^m\right)^n=a^{m n}\)
Proof:
By definition we get,
\(\left(a^m\right)^n=a^m \times a^m \times a^m \times\)…………….(n-times) [by definition]
\(a^{m+m+m+}(n-times)\left[\right.by formula-1]=a^{m n}\)∴ \(\left(a^m\right)^n=a^{m n}\) proved
Formula 4: \((a b)^m=a^m \cdot b^m\)
Proof:
By definition we get,
\((a b)^m=a b \times a b \times a b \times\)……………..(m times)[by definition]
\(\{a \times a \times a \times \ldots \ldots .(m \text { times })\} \times\{b \times b \times b \times \ldots \ldots .(m \text { times })\}\)\(a^{1+1+1+}(m times) \times b^{1+1+1+}\)………..(m times)[by formula – 1]
= \(a^m \times b^m\)
∴ \((a b)^m=a^m \cdot b^m\)
Formula 5: \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}, b \neq 0\)
Proof:
By definition we get,
\(\left(\frac{a}{b}\right)^m=\frac{a}{b} \times \frac{a}{b} \times \frac{a}{b} \times\)…………….(m times) [by definition]
\(\frac{a \times a \times a \times \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots(m \text { times })}{b \times b \times b \times \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(m \text { times })}\) \(\frac{a^{1+1+1+\ldots . \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .(m \text { times })}}{b^{1+1+1+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .(m \text { times })}}=\frac{a^m}{b^m}\)∴ \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}, b \neq 0\) proved
From the above laws we get the following corollaries:
Corollary 1:
If l, m, n be positive integers, then \(a^l \times a^m \times a^n=a^{l+m+n}\)
In general, \(\begin{aligned}
&a^{n_1} \times a^{n_2} \times a^{n_3} \times\\
&. . \times a^{n_n}=a^{n_{\mathrm{r}}+n_2+n_3+\ldots \ldots \ldots \ldots+n_n}
\end{aligned}\)
Corollary 2:
If l, m, n be positive integers, then \(\left(\left(a^l\right)^m\right)^n=\left(a^{l m}\right)^n=a^{l m n}\)
In general, \(\left(\left(a^{n_1}\right)^{n_2 \ldots \ldots \ldots \ldots}\right)^{n_n}=a^{n_1 n_2 n_3 \ldots \ldots \ldots \ldots \ldots n_n}\)
Corollary 3:
If m be a positive integer, then \((a b c)^m=a^m \cdot b^m \cdot c^m\)
In general, \(\left(a_1 a_2 a_3 \ldots \ldots \ldots . a_n\right)^m=a_1^m \cdot a_2^m \cdot a_3^m \ldots \ldots \ldots \ldots a_n^m \text {. }\)
Corollary 4: If m be any positive integer, then a \(a^{-m}=\frac{1}{a^m}\)
Proof:
From formula-1 we get if m and n be any two positive integers, then \(a^m \times a^n=a^{m+n}\)………….(1)
Putting n = -m in (1) we get, \(a^m \times a^{-m}=a^{m-m}\)
∴ \(a^{-m}=\frac{1}{a^m}\) (Proved).
Corollary-5:
If \(a_1, a_2 a_3,……………… a_n and b_1, b_2, b_3,………….b_n\), be any real number and m be any positive integer, then
\(\left(\frac{a_1+a_2+a_3+\ldots \ldots \ldots \ldots \ldots+a_n}{b_1+b_2+b_3+\ldots \ldots \ldots \ldots \ldots+b_n}\right)^m=\frac{\left(a_1+a_2+a_3+\ldots \ldots \ldots+a_n\right)^m}{\left(b_1+b_2+b_3+\ldots \ldots \ldots \ldots \ldots+b_n\right)^m}\)
Corollary 6:
If a be any real number and m and n be any positive integer (n + 0), then
1. \(\sqrt[n]{a}=a^{\frac{1}{n}}\)
2. \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\)
Some proofs:
1. Prove that a° = 1 (a = 0).
Proof: We know, \(\frac{a^m}{a^n}=a^{m-n}\)……………………..(1) [by formula-2]
where a is any real number and m, and n are positive integers.
Now, putting n instead of m in (1) we get, \(\frac{a^n}{a^n}=a^{n-n}\)
or, \(1=a^{n-n}\)
or, 1 = a° [ ∵ n – n = 0]
∴ a° = 1 (Proved)
2. Prove that if ax = 1, (where a > 0 and a not = 1), then x = 0.
Proof:
Let x > 0,
∴ ax = 1 when a = 1, since 1x = 1 for x > 0.
But given that a ≠ 1,
∴ x <≠ 0.
Now, if x = 0, then ax = a° = 1,
∴ if ax = 1, then x = 0 (Proved).
3. Prove that if a = 1, (where a > 0 and x not = 1), then a = 1.
Proof:
Given a = 1, (where a > 0 and x not = 1)
If possible, let a ≠ 1.
Then by the proof described in 2 above, we get, x = 0.
But given that x ≠ 0.
∴ a = 1 (Proved).
4. Prove that if a, b, x be any real numbers and if ax = ay (where a not = 0 or ± 1 or ±∞), then x = y.
Proof:
Given that ax = ay
or, \(\frac{a^x}{a^y}=1\) =1 [ ∵ a not = 0 or, ± 1 or, ± ∞]
or, ax-y = 1 [by Formula-2]
or, ax-y = a° [by the proof-1]
∴ x – y = 0 or, x = y (Proved).
5. Prove that if a, b, x be any real number and a = b (b ≠ 0), then either a = b or x = 0.
Proof:
Given that ax = bx or, \(\frac{a^x}{a^x}=1\) [∵ b ≠ 0]
or, \(\left(\frac{a}{b}\right)^x=1\)
or, \(\left(\frac{a}{b}\right)^x=\left(\frac{a}{b}\right)^0\) [∵ \(\left(\frac{a}{b}\right)^0=1\) ]
Now, if x ≠ 0, then \(\frac{a}{b}\) = 1 [Since 1x = 1°]
or, a = b.
Again, if \(\frac{a}{b}\) ≠ 1, then x = 0.
∴ either a = b or x = 0 (Proved)
Observe the following examples minutely to clear your conceptions about the application of laws of indices.
Chapter 6 Laws Of Indices Select The Correct Answer (MCQ)
Question 1.
1. The value of \((0 \cdot 243)^{0 \cdot 2} \times(10)^{0 \cdot 6}\) is
- 0.3
- 3
- 0.9
- 9
Solution:
\((0 \cdot 243)^{0 \cdot 2} \times(10)^{0 \cdot 6}\)
= \((0.243)^{\frac{2}{10}} \times(10)^{\frac{6}{10}}=(0.243)^{\frac{1}{5}} \times(10)^{\frac{3}{5}}\)
=\(\left(\frac{243}{1000}\right)^{\frac{1}{5}} \times(10)^{\frac{3}{5}}\)
= \((243)^{\frac{1}{5}} \times\left(\frac{1}{1000}\right)^{\frac{1}{5}} \times(10)^{\frac{3}{5}}\)
= \(\left(3^5\right)^{\frac{1}{5}} \times\left\{\frac{1}{(10)^3}\right\}^{\frac{1}{5}} \times(10)^{\frac{3}{5}}\)
= \(3^5 \times \frac{1}{5} \times\left(10^{-3}\right)^{\frac{1}{5}} \times(10)^{\frac{3}{5}}\)
= \(3^1 \times 10^{-\frac{3}{5}} \times 10^{\frac{3}{5}}=3 \times 10^{-\frac{3}{5}+\frac{3}{5}}\)
= 3 x (10)º
= 3 x 1
= 3.
2. The value of \(2^{\frac{1}{2}} \times 2^{-\frac{1}{2}} \times(16)^{\frac{1}{2}}\) is
- 1
- 2
- 4
- 1/2
Solution:
\(2^{\frac{1}{2}} \times 2^{-\frac{1}{2}} \times(16)^{\frac{1}{2}}\)= \(2^{\frac{1}{2}-\frac{1}{2}} \times\left(4^2\right)^{\frac{1}{2}}\)
= \(2^0 \times 4^{2 \times \frac{1}{2}}\)
=1 x 4
=4
3. If 4x = 8³, then x =
- 3/2
- 9/2
- 3
- 9
Solution :
4x = 8³
or, \(\left(2^2\right)^x=\left(2^3\right)^3\)
or, \(2^{2 x}=2^9\)
⇒ 2x = 9
x = \(\frac{9}{2}\)
4. If 20-x = \(\frac{1}{7}\), then (20)2x =
- 1/49
- 7
- 49
- 1
Solution:
20-x = \(\frac{1}{7}\)
or, \(\frac{1}{20^x}=\frac{1}{7} \text { or, } 20^x=7 \text { or, }\left(20^x\right)^2=7^2 \text { or, }(20)^{2 x}=49\)
5. If 4 x 5x = 500, then xx =
- 8
- 1
- 64
- 27
Solution:
4 x 5x = 500
or, 5x = \(\frac{500}{4}\)
or, 5x = 125
or, 5x = 5³
or, x = 3
∴ xx = 3³
= 27
6. \(\left\{(81)^{-\frac{3}{4}} \times \frac{(16)^{\frac{1}{4}}}{6^{-2}} \times\left(\frac{1}{27}\right)^{-\frac{4}{3}}\right\}^{\frac{1}{3}}\) = 1
- 4
- 5
- 6
- 8
Solution:
\(\left\{(81)^{-\frac{3}{4}} \times \frac{(16)^{\frac{1}{4}}}{6^{-2}} \times\left(\frac{1}{27}\right)^{-\frac{4}{3}}\right\}^{\frac{1}{3}}\)= \(\left\{(3)^{4 \times-\frac{3}{4}} \times \frac{(2)^{4 \times \frac{1}{4}}}{\frac{1}{36}} \times(3)^{-3 \times-\frac{4}{3}}\right\}^{\frac{1}{3}}\)
= \(\left\{3^{-3} \times \frac{2}{\frac{1}{36}} \times(3)^4\right\}^{\frac{1}{3}}=\left\{(3)^{-3+4} \times 2 \times 36\right\}^{\frac{1}{3}}=(216)^{\frac{1}{3}}=\left(6^3\right)^{\frac{1}{3}}=6^{3 \times \frac{1}{3}}\)
= 6¹
= 6.
7. \(\sqrt[4]{x^{\frac{17}{2}} y^{\frac{5}{2}} \sqrt{x^{\frac{5}{2}} \sqrt{x^{-7} y^6}}}\) =
- xy
- x²y
- xy²
- x²y²
Solution:
\(\sqrt[4]{x^{\frac{17}{2}} y^{\frac{5}{2}} \sqrt{x^{\frac{5}{2}} \sqrt{x^{-7} y^6}}}\)= \(\sqrt[4]{x^{\frac{17}{2}} y^{\frac{5}{2}} \sqrt{x^{-1} \cdot y^3}}=\sqrt[4]{x^{\frac{17}{2}} y^{\frac{5}{2}} \cdot x^{-\frac{1}{2}} \cdot y^{\frac{3}{2}}}=\sqrt[4]{x^8 y^4}\)
= x²y
8. \((\sqrt[5]{8})^{\frac{5}{2}} \times(16)^{-\frac{3}{2}}\) =
- \(2^{-\frac{9}{2}}\)
- \(2^{\frac{9}{2}}\)
- \(2^{-\frac{7}{2}}\)
- \(2^{\frac{7}{2}}\)
Solution:
\((\sqrt[5]{8})^{\frac{5}{2}} \times(16)^{-\frac{3}{2}}\)= \(8^{\frac{1}{5} \times \frac{5}{2}^5} \times(16)^{-\frac{3}{2}}\)
= \(\left(2^3\right)^{\frac{1}{2}} \times\left(2^4\right)^{-\frac{3}{2}}=2^{3 \times \frac{1}{2}} \times 2^{4 \times-\frac{3}{2}}=2^{\frac{3}{2}} \times 2^{-6}=2^{\frac{3}{2}-6}=2^{-\frac{9}{2}}\)
9. \(4^{\frac{1}{3}} \times\left[2^{\frac{1}{3}} \times 3^{\frac{1}{2}}\right] \div 9^{\frac{1}{4}}\)
- 1
- 2
- -2
- 4
Solution:
\(4^{\frac{1}{3}} \times\left[2^{\frac{1}{3}} \times 3^{\frac{1}{2}}\right] \div 9^{\frac{1}{4}}\)= \(\left(2^2\right)^{\frac{1}{3}} \times 2^{\frac{1}{3}} \times 3^{\frac{1}{2}} \div\left(3^2\right)^{\frac{1}{4}}\)
= \(2^{\frac{2}{3}} \times 2^{\frac{1}{3}} \times 3^{\frac{1}{2}} \div 3^{\frac{1}{2}}=2^{\frac{2}{3}+\frac{1}{3}} \times 3^{\frac{1}{2}-\frac{1}{2}}=2^1 \times 3^0\)
= 2 x 1 [∵ 3º = 1]
= 2
10. \(\left\{(125)^{-2} \times(16)^{-\frac{3}{2}}\right\}^{-\frac{1}{6}}\) =
- 4
- 8
- 10
- 11
Solution:
\(\left\{(125)^{-2} \times(16)^{-\frac{3}{2}}\right\}^{-\frac{1}{6}}\)= \(\left\{\left(5^3\right)^{-2} \times\left(2^4\right)^{-\frac{3}{2}}\right\}^{-\frac{1}{6}}\)
= \(\left\{5^{-6} \times 2^{4 \times-\frac{3}{2}}\right\}^{-\frac{1}{6}}=\left(5^{-6} \times 2^{-6}\right)^{-\frac{1}{6}}=5^{-6 \times-\frac{1}{6}} \times 2^{-6 \times-\frac{1}{6}}=5^1 \times 2^1\)
= 5 x 2
= 10
Chapter 6 Laws Of Indices Short Answer Type Questions
Question 1.
1. If \((27)^x=(81)^y\), then find x: y,
Solution:
\((27)^x=(81)^y\)or, \(\left(3^3\right)^x=\left(3^4\right)^y \text { or, } 3^{3 x}=3^{4 y} \text { or, } 3 x=4 y \text { or, } \frac{x}{y}=\frac{4}{3}\)
∴ x: y = 4: 3.
2. Which one of the two numbers \(3^{3^3} \text { and }\left(3^3\right)^3\) is greater?
Solution:
\(3^{3^3}=3^{27}=\left(3^3\right)^9=(27)^9\)Again, \(\left(3^3\right)^3=(27)^3\)
∴ \(3^{3^3}>\left(3^3\right)^3\) [ ∵ \((27)^9>(27)^3\) ]
∴ \(3^{3^3}\) is greater.
3. Find the value of \(\sqrt[3]{\left(\frac{1}{64}\right)^{\frac{1}{2}}}\)
Solution:
\(\sqrt[3]{\left(\frac{1}{64}\right)^{\frac{1}{2}}}\)= \(\sqrt[3]{\left(\frac{1}{2^6}\right)^{\frac{1}{2}}}=\sqrt[3]{\left(2^{-6}\right)^{\frac{1}{2}}}=\sqrt[3]{2^{-6 \times \frac{1}{2}}}=\sqrt[3]{2^{-3}}=\left(2^{-3}\right)^{\frac{1}{3}}=2^{-3 \times \frac{1}{3}}=2^{-1}\)
= \(\frac{1}{2}\)
\(\sqrt[3]{\left(\frac{1}{64}\right)^{\frac{1}{2}}}\) = \(\frac{1}{2}\)
4. \(\left(5^5+0.01\right)^2-\left(5^5-0.01\right)^2=5^x\) then what is the value of x?
Solution:
\(\left(5^5+0.01\right)^2-\left(5^5-0.01\right)^2=5^x\)or, \(\left(5^5+0.01+5^5-0.01\right)\left(5^5+0.01-5^5+0.01\right)=5^x\)
or, \(\left(2.5^5\right) \times(0.02)=5^x \text { or, } 0.04 \times 5^5=5^x \text { or, } \frac{4}{100} \times 5^5=5^x \text { or, } \frac{1}{25} \times 5^5=5^x\)
or, \(\frac{1}{5^2} \times 5^5=5^x \text { or, } 5^{-2} \times 5^5=5^x \text { or, } 5^{-2+5}=5^x \text { or, } 5^3=5^x\)
or, x = 3.
The value of x = 3.
5.If \(3 \times 27^x=9^{x+4}\), then find the value of x.
Solution:
\(3 \times 27^x=9^{x+4}\)or, \(3 \times\left(3^3\right)^x=\left(3^2\right)^{x+4} \text { or, } 3 \times 3^{3 x}=3^{2(x+4)} \text { or, } 3^{1+3 x}=3^{2 x+8}\)
∴ 1 + 3x = 2x + 8
or, 3x – 2x = 8 – 1
or, x = 7.
The value of x = 7.
6. Find the value of \(\frac{\frac{1}{4^{-3}}-\frac{2}{10^{-2}}}{\frac{1}{2^{-2}}+\frac{1}{4^{-1}}}\)
Solution:
\(\frac{\frac{1}{4^{-3}}-\frac{2}{10^{-2}}}{\frac{1}{2^{-2}}+\frac{1}{4^{-1}}}\)= \(\frac{4^3-2 \times 10^2}{2^2+4}=\frac{64-200}{4+4}=\frac{-136}{8}\)
= -17.
\(\frac{\frac{1}{4^{-3}}-\frac{2}{10^{-2}}}{\frac{1}{2^{-2}}+\frac{1}{4^{-1}}}\) = -17.
7. Find the value of \(\sqrt[l n]{\frac{x^l}{x^n}} \times \sqrt[m n]{\frac{x^n}{x^m}} \times \sqrt[l n]{\frac{x^m}{x^l}}\)
Solution:
\(\sqrt[l n]{\frac{x^l}{x^n}} \times \sqrt[m n]{\frac{x^n}{x^m}} \times \sqrt[l n]{\frac{x^m}{x^l}}\)= \(\left(x^{l-n}\right)^{\frac{1}{l n}} \times\left(x^{n-m}\right)^{\frac{1}{m n}} \times\left(x^{m-l}\right)^{\frac{1}{l m}}\)
= \((x)^{\frac{1}{n}-\frac{1}{l}} \times(x)^{\frac{1}{m}-\frac{1}{n}} \times(x)^{\frac{1}{l}-\frac{1}{m}}=x^{\frac{1}{n}-\frac{1}{l}+\frac{1}{m}-\frac{1}{n}+\frac{1}{l}-\frac{1}{m}}\)
= xº
= 1.
8. Calculate: \(\left\{\frac{\left(9^{n+\frac{1}{4}}\right) \sqrt{3.3^n}}{3 \sqrt{3^{-n}}}\right\}^{\frac{1}{n}}\)
Solution:
\(\left\{\frac{\left(9^{n+\frac{1}{4}}\right) \sqrt{3.3^n}}{3 \sqrt{3^{-n}}}\right\}^{\frac{1}{n}}\)= \(\left[\frac{\left\{\left(3^2\right)^{n+\frac{1}{4}}\right\} 3^{\frac{1}{2}} \cdot 3^{\frac{n}{2}}}{3.3^{-\frac{n}{2}}}\right]^{\frac{1}{n}}\)
= \(\left(\frac{3^{2 n+\frac{1}{2}} \cdot 3^{\frac{1}{2}} \cdot 3^{\frac{n}{2}}}{3.3^{-\frac{n}{2}}}\right)^{\frac{1}{n}}\)
= \(\left(\frac{3^{2 n+\frac{1}{2}+\frac{1}{2}+\frac{n}{2}}}{3^{1-\frac{n}{2}}}\right)^{\frac{1}{n}}\)
= \(\left\{\frac{(3)^{\frac{5 n+2}{2}}}{(3)^{\frac{2-n}{2}}}\right\}^{\frac{1}{n}}\)
= \(\left\{(3) \frac{5 n+2}{2}-\frac{2-n}{2}\right\}^{\frac{1}{n}}\)
= \(\left\{(3)^{\frac{5 n+2-2+n}{2}}\right\}^{\frac{1}{n}}=\left(3^{3 n}\right)^{\frac{1}{n}}=3^{3 n \times \frac{1}{n}}\)
= 3³
= 27
9. Simplify: \(\frac{x^{a+b} \cdot x^{a-b} \cdot x^{c-2 a}}{x^{c-a}}\)
Solution:
\(\frac{x^{a+b} \cdot x^{a-b} \cdot x^{c-2 a}}{x^{c-a}}\)= \(\frac{x^{a+b+a-b+c-2 a}}{x^{c-a}}=\frac{x^c}{x^{c-a}}=x^{c-c+a}=x^a\)
10. Simplify: \(\left\{\left(x^{a+b-c} \times x^{a-b+c}\right)^b\right\}^c\)
Solution:
\(\left\{\left(x^{a+b-c} \times x^{a-b+c}\right)^b\right\}^c\)= \(\left\{\left(x^{a+b-c+a-b+c}\right)^b\right\}^c=\left\{\left(x^{2 a}\right)^b\right\}^c=\left(x^{2 a b}\right)^c=x^{2 a b c}\)
Chapter 6 Laws Of Indices Long Answer Type Questions
Question 1.
1. \(\left(\frac{4^{m+\frac{1}{4}} \times \sqrt{2.2^m}}{2 . \sqrt{2^{-m}}}\right)^{\frac{1}{m}}\)
Solution:
\(\left(\frac{4^{m+\frac{1}{4}} \times \sqrt{2.2^m}}{2 . \sqrt{2^{-m}}}\right)^{\frac{1}{m}}\)= \(\left\{\frac{\left(2^2\right)^{m+\frac{1}{4}} \times 2^{\frac{1}{2}} \cdot 2^{\frac{m}{2}}}{2.2^{-\frac{m}{2}}}\right\}^{\frac{1}{m}}=\left(\frac{2^{2 m+\frac{1}{2}} \times 2^{\frac{1}{2}+\frac{m}{2}}}{2^{1-\frac{m}{2}}}\right)^{\frac{1}{m}}\)
= \(\left\{\frac{(2)^{2 m+\frac{1}{2}+\frac{1}{2}+\frac{m}{2}}}{(2)^{1-\frac{m}{2}}}\right\}^{\frac{1}{m}}=\left\{\frac{(2)^{\frac{5 m+2}{2}}}{(2)^{\frac{2-m}{2}}}\right\}^{\frac{1}{m}}=\left\{(2)^{\frac{5 m+2}{2}-\frac{2-m}{2}}\right\}^{\frac{1}{m}} .\)
= \(\left\{(2)^{\frac{5 m+2-2+m}{2}}\right\}^{\frac{1}{m}}=\left(2^{3 m}\right)^{\frac{1}{m}}=2^{3 m \times \frac{1}{m}}=2^3\)
= 8
2. \(\left(\frac{x^a}{x^b}\right)^{a^2+a b+b^2} \times\left(\frac{x^b}{x^c}\right)^{b^2+b c+c^2} \times\left(\frac{x^c}{x^a}\right)^{c^2+c a+a^2}\)
Solution:
\(\left(\frac{x^a}{x^b}\right)^{a^2+a b+b^2} \times\left(\frac{x^b}{x^c}\right)^{b^2+b c+c^2} \times\left(\frac{x^c}{x^a}\right)^{c^2+c a+a^2}\)= \(x^{(a-b)\left(a^2+a b+b^2\right)} \times x^{(b-c)\left(b^2+b c+c^2\right)} \times x^{(c-a)\left(c^2+c a+a^2\right)}\)
= \(x^{a^3-b^3} \times x^{b^3-c^3} \times x^{c^3-a^3}=x^{a^3-b^3+b^3-c^3+c^3-a^3}\)
= xº
= 1.
Question 2.
1. \(5^{\frac{1}{2}}, 10^{-\frac{1}{4}}, 6^{\frac{1}{3}}\)
Solution:
\(5^{\frac{1}{2}}, 10^{-\frac{1}{4}}, 6^{\frac{1}{3}}\)Now, \(5^{\frac{1}{2}}=5^{6 \times \frac{1}{12}}=\left(5^6\right)^{\frac{1}{12}}=(15625)^{\frac{1}{12}}\)
\((10)^{-\frac{1}{4}}=10^{-3 \times \frac{1}{12}}=\left(10^{-3}\right)^{\frac{1}{12}}=\left(\frac{1}{1000}\right)^{\frac{1}{12}}\) \(6^{\frac{1}{3}}=6^{4 \times \frac{1}{12}}=\left(6^4\right)^{\frac{1}{12}}=(1296)^{\frac{1}{12}}\)Noe, \(\frac{1}{1000}<1296<15625\)
∴ \(\left(\frac{1}{1000}\right)^{\frac{1}{12}}<(1296)^{\frac{1}{12}}<(15625)^{\frac{1}{12}}\)
or, \((10)^{-\frac{1}{4}}<6^{\frac{1}{3}}<5^{\frac{1}{2}}\)
∴ rearranging according to the ascending values we get, \(10^{-\frac{1}{4}}, 6^{\frac{1}{3}}, 5^{\frac{1}{2}}\)
2. \(2^{60}, 3^{48}, 4^{36}, 5^{24}\)
Solution:
\(2^{60}, 3^{48}, 4^{36}, 5^{24}\)Now, \(2^{60}=2^{5 \times 12}=\left(2^5\right)^{12}=(32)^{12} ; 3^{48}=3^{4 \times 12}=\left(3^4\right)^{12}=(81)^{12}\)
\(3^{36}=3^{3 \times 12}=\left(3^3\right)^{12}=(27)^{12} ; 5^{24}=5^{2 \times 12}=\left(5^2\right)^{12}=(25)^{12}\)Then \(25<27<32<81 \text { or, }(25)^{12}<(27)^{12}<(32)^{12}<(81)^{12}\)
∴ rearranging according to the ascending values we get, \(5^{24}, 3^{36}, 2^{60}, 3^{48}\)
Question 3. Prove that:
1. \(\left(\frac{a^q}{a^r}\right)^p \times\left(\frac{a^r}{a^p}\right)^q \times\left(\frac{a^p}{a^q}\right)^r=1\)
Solution:
Proof:
LHS = \(\left(\frac{a^q}{a^r}\right)^p \times\left(\frac{a^r}{a^p}\right)^q \times\left(\frac{a^p}{a^q}\right)^r\)
= \(a^{p q-p r} \times a^{q r-p q} \times a^{p r-q r}=a^{p q-p r+q r-p q+p r-q r}\)
= aº
= 1. (proved)
2. \(\left(\frac{x^m}{x^n}\right)^{m+n} \times\left(\frac{x^n}{x^l}\right)^{n+l} \times\left(\frac{x^l}{x^m}\right)^{l+m}=1\)
Solution:
Proof:
LHS \(\left(\frac{x^m}{x^n}\right)^{m+n} \times\left(\frac{x^n}{x^l}\right)^{n+l} \times\left(\frac{x^l}{x^m}\right)^{l+m}\)
= \(=x^{(m-n)(m+n)} \times x^{(n-l)(n+l)} \times x^{(l-m)(l+m)}=x^{m^2-n^2} \times x^{n^2-l^2} \times x^{l^2-m^2}\)
= \(x^{m^2-n^2+n^2-l^2+l^2-m^2}\)
= xº
= 1. (proved)
3. \(\left(\frac{x^m}{x^n}\right)^{m+n-l} \times\left(\frac{x^n}{x^l}\right)^{n+l-m} \times\left(\frac{x^l}{x^m}\right)^{l+m-n}\)
Solution:
Proof:
LHS \(\left(\frac{x^m}{x^n}\right)^{m+n-l} \times\left(\frac{x^n}{x^l}\right)^{n+l-m} \times\left(\frac{x^l}{x^m}\right)^{l+m-n}\)
= \(\left(x^{m-n}\right)^{m+n-l} \times\left(x^{n-l}\right)^{n+l-m} \times\left(x^{l-m}\right)^{l+m-n}\)
= \(x^{(m-n)(m+n-l)} \times x^{(n-l)(n+l-m)} \times x^{(l-m)(l+m-n)}\)
= \(x^{(m-n)(m+n)-l(m-n)} \times x^{(n-l)(n+l)-m(n-l)} \times x^{(l-m)(l+m)-n(l-m)}\)
= \(x^{m^2-n^2-l m+l n} \times x^{n^2-l^2-m n+l m} \times x^{l^2-m^2-n l+m n}\)
= \(x^{m^2-n^2-l m+l n+n^2-l^2-m n+l m+l^2-m^2-l n+m n}\)
= xº
= 1. (proved)
4.\(\left(a^{\frac{1}{x-y}}\right)^{\frac{1}{x-z}} \times\left(a^{\frac{1}{y-z}}\right)^{\frac{1}{y-x}} \times\left(a^{\frac{1}{z-x}}\right)^{\frac{1}{z-y}}\)
Solution:
LHS \((a)^{\frac{1}{(x-y)(x-z)}} \times a^{\frac{1}{(y-z)(y-x)}} \times a^{\frac{1}{(z-x)(z-y)}}\)
= \((a)^{\frac{1}{-(x-y)(z-x)}} \times a^{\frac{1}{-(y-z)(x-y)}} \times a^{\frac{1}{-(z-x)(y-z)}}\)
= \((a)^{\frac{1}{-(x-y)(z-x)}}+\frac{1}{-(y-z)(x-y)}+\frac{1}{-(z-x)(y-z)}\)
= \((a)^{\frac{y-z+z-x+x-y}{-(x-y)(y-z)(z-x)}}=(a)^{\frac{0}{-(x-y)(y-z)(z-x)}}\)
= (a)º
= 1. (proved)
Question 4. If x + x = 2y and b² = ac, then prove that \(a^{y-z} \cdot b^{z-x} \cdot c^{x-y}=1\)
Solution:
Given that x + z =2y
or, x – y = y – z
∴ LHS = \(a^{y-z} \cdot b^{z-x} \cdot c^{x-y}=a^{x-y} \cdot b^{z-x} \cdot c^{x-y_1}\) [ ∵ y – z = x – y]
= \((a c)^{x-y} \cdot b^{z-x}=\left(b^2\right)^{x-y} \cdot b^{z-x}\) [∵ b² = ac ]
= \(b^{2 x-2 y} \cdot b^{z-x}=b^{2 x-2 y+z-x}=b^{x-2 y+z}\)
= bº [ ∵ x + z = 2y or, x – 2y + z = 0]
= 1 (proved)
Question 5. If \(a=x y^{p-1}, \quad b=x y^{q-1}, \quad c=x y^{r-1}\), then show that \(a^{q-r} \cdot b^{r-p} \cdot c^{p-q}=1\)
Solution:
Given that \(a=x y^{p-1}, \quad b=x y^{q-1}, \quad c=x y^{r-1}\)
∴ LHS = \(a^{q-r} \cdot b^{r-p} \cdot c^{p-q}=\left(x y^{p-1}\right)^{q-r} \cdot\left(x y^{q-1}\right)^{r-p}\left(x y^{r-1}\right)^{p-q}\)
= \((x)^{q-r} \cdot(y)^{(p-1)(q-r)} \cdot(x)^{r-p} \cdot(y)^{(q-1)(r-p)} \cdot(x)^{p-q} \cdot(y)^{(r-1)(p-q)}\)
= \((x)^{q-r+r-p+p-q} \cdot(y)^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}\)
= \((x)^0 \cdot(y)^{p q-p r-q+r+q r-p q-r+p+p r-q r-p+q}\)
= 1. yº
= 1.1
= 1 (proved)
Question 6. If \(x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}\) and xyz =1, then show that a+b+c = 0.
Solution:
\(x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}\) = p (let)
∴ \(x^{\frac{1}{a}}=p \Rightarrow x=p^a ; \quad y^{\frac{1}{b}}=p \Rightarrow y=p^b ; z^{\frac{1}{c}}=p \Rightarrow z=p^c\)
Now, given that xyz = 1
or, \(p^a \cdot p^b \cdot p^c=1 \quad \text { or, } p^{a+b+c}=1=p^{\circ}\) [∵ pº = 1]
∴ a+b+c = 0 (proved)
Question 7. If \(a^x=b^y=c^z\) and abc = 1, then prove that xy + yz + zx = 0
Solution:
Given that \(a^x=b^y=c^z\) = r (let)
∴ \(a^x=r \Rightarrow a=r^{\frac{1}{x}} ; b^y=r \Rightarrow b=r^{\frac{1}{y}} ; c^z=r \Rightarrow c=r^{\frac{1}{z}}\)
Also, given that abc = 1
or, \(r^{\frac{1}{x}} \cdot r^{\frac{1}{y}} \cdot r^{\frac{1}{z}}=1 \text { or, } r^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)
= 1
= rº [∵ rº = 1]
∴ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \text { or, } \frac{y z+z x+x y}{x y z}=0 \text { or, } x y+y z+z x=0\) (Proved)
Question 8. Solve:
1. \(2^{x+2}+2^{x-1}=9\)
Solution:
\(2^{x+2}+2^{x-1}=9\)or, \(2^x \cdot 2^2+2^x \cdot 2^{-1}=9 \text { or, } 2^x\left(4+\frac{1}{2}\right)=9\)
or, \(2^x \times \frac{9}{2}=9 \text { or, } 2^x=2^1\)
∴ x = 1
∴ the required solution : x = 1
2. \(2^{4 x} \cdot 4^{3 x-1}=\frac{4^{2 x}}{2^{3 x}}\)
Solution:
\(2^{4 x} \cdot 4^{3 x-1}=\frac{4^{2 x}}{2^{3 x}} \text { or, } 2^{4 x} \cdot\left\{(2)^2\right\}^{3 x-1}=\frac{\left(2^2\right)^{2 x}}{2^{3 x}}\)or, \(2^{4 x} \cdot 2^{6 x-2}=\frac{2^{4 x}}{2^{3 x}}\)
or, \(2^{4 x+6 x-2}=2^{4 x-3 x} \text { or, } 2^{10 x-2}=2^x \Rightarrow 10 x-2=x \text { or, } 9 x=2\)
or, \(x=\frac{2}{9}\)
∴ the required solution : x = \(\frac{2}{9}\)
3. \(9 \times 81^x=27^{2-x}\)
Solution:
\(9 \times(81)^x=27^{2-x} \text { or, } 3^2 \times\left(3^4\right)^x=\left(3^3\right)^{2-x}\)or, \(3^2 \times 3^{4 x}=3^{3(2-x)} \text { or, } 3^{2+4 x}=3^{6-3 x}\)
∴ 2 + 4x = 6 – 3x
or, 4x + 3x = 6-2
or, 7x = 4
or, x = \(\frac{4}{7}\)
∴ the required solution: x = \(\frac{4}{7}\)
4. \(6^{2 x+4}=3^{3 x} \cdot 2^{x+8}\)
Solution:
\(6^{2 x+4}=3^{3 x} \cdot 2^{x+8} \text { or }(2 \times 3)^{2 x+4}=3^{3 x} \cdot 2^{x+8}\)or, \(2^{2 x+4} \cdot 3^{2 x+4}=3^{3 x} \cdot 2^{x+8}\)
or, \(\frac{3^{2 x+4}}{3^{3 x}}=\frac{2^{x+8}}{2^{2 x+4}} \text { or, } 3^{2 x+4-3 x}=2^{x+8-2 x-4} \text { or, } 3^{4-x}=2^{4-x}\)
or, \(\frac{3^{4-x}}{2^{4-\dot{x}}}=1 \text { or, }\left(\frac{3}{2}\right)^{4-x}=\left(\frac{3}{2}\right)^0\) [∵ \(\left(\frac{3}{2}\right)^0=1\)]
or, 4 – x = 0
or, x = 4
∴ The required solution: x = 4.
