WBBSE Solutions For Class 9 Maths Algebra Chapter 6 Laws Of Indices

WBBSE Solutions For Class 9 Maths 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

1. aⁿ is not called the index. It is a representation in the form of an index.
2. n is called the index.
3. a is called the base.
4. Instead of a positive integer, if n is a zero, then an = 1 when a 0.
5. If n be a negative integer, then a not = a x a x a x a x ……(n-times).
6. Then \(a^n=a^{-m}(\text { where } n=-m, m>0)=\frac{1}{a^m}\)
   = \(\frac{1}{a \times a \times a \times \ldots \ldots(m-\text { times })}\)
7. If n is any positive rational number, let where n = \(\frac{p}{q}\), both p and q are positive integers, and q ≠  0.
   
   ∴ \(a^n=a^{\frac{p}{q}}=\sqrt[q]{a \times a \times a \times \ldots \ldots \ldots(p-\text { times })}\)
8. If n is any negative rational number, let n = – \(\frac{p}{q}\) where both p and q are positive integers and p ≠ 0.\(a^n=a^{-\frac{p}{q}}=\frac{1}{\sqrt{a^p}}=\frac{1}{\sqrt[2]{a \times a \times a \times 1 \ldots. .(p-\text { times })}}\)
9. If n is any irrational number, let n√p, where p > 0 and p is not a perfect square. \(a^n=a^{\sqrt{p}}\)

WBBSE Class 9 Maths 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.

WBBSE Class 9 Maths 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 = x^1/2

If n = 3, then a3 = x 

or, a=3√x 

or, a=x^1/3

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 a^1/2? The answer is the meaning of a^1/2 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, 4^1/2 = (+2), since (+ 2) x (+ 2) = 4.

Again, 4 (-2), since (-2) × (-2) = 4.

∴ We can say that 4² is not meaningless and in fact 4^1/2 = ±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 a^1/3 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.

WBBSE Class 9 Maths Chapter 6 Laws Of Indices Laws Of Indices

WBBSE Class 9 Laws of Indices Overview

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

Understanding Laws of Indices for Class 9

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)

Examples of Laws of Indices in Algebra

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

Practice Problems on Laws of Indices

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. \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\)

Common Mistakes in Using Indices

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, 

∴ a^x = 1 when a = 1, since 1^x = 1 for x > 0.

But given that a ≠ 1,   

∴ x <≠ 0.

Now, if x = 0, then a^x = a° = 1, 

∴ if a^x = 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 a^x = a^y

or, \(\frac{a^x}{a^y}=1\) =1 [ ∵ a not = 0 or, ± 1 or, ± ∞]

or, a^x-y = 1 [by Formula-2]

or, a^x-y = a° [by the proof-1]

∴ x – y = 0 or, x = y (Proved).

WBBSE Class 9 Revision Notes on Indices

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 a^x = b^x 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 1^x =  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)

Step-by-Step Solutions for Laws of Indices

Question 1. The value of \((0 \cdot 243)^{0 \cdot 2} \times(10)^{0 \cdot 6}\) is

1. 0.3
2. 3
3. 0.9
4. 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.

WBBSE Class 9 Algebra Solutions Question 2. The value of  \(2^{\frac{1}{2}} \times 2^{-\frac{1}{2}} \times(16)^{\frac{1}{2}}\) is

1. 1
2. 2
3. 4
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

Question 3. If 4^x = 8³, then x = 

1. 3/2
2. 9/2
3. 3
4. 9

Solution :

4^x = 8³

or, \(\left(2^2\right)^x=\left(2^3\right)^3\)

or, \(2^{2 x}=2^9\)

⇒ 2x = 9

x = \(\frac{9}{2}\)

Question 4. If 20^-x = \(\frac{1}{7}\), then (20)^2x =

1. 1/49
2. 7
3. 49
4. 1

Solution: 

20^-x = \(\frac{1}{7}\)

or, \(\frac{1}{20^x}=\frac{1}{7} \text { or, } 20^x=7 \text { or, }\left(20^x\right)^2=7^2 \text { or, }(20)^{2 x}=49\)

Question 5. If 4 x 5^x = 500, then x^x =

1. 8
2. 1
3. 64
4. 27

Solution:

4 x 5^x = 500

or, 5^x = \(\frac{500}{4}\)

or, 5^x = 125

or, 5^x = 5³

or, x = 3

∴ x^x = 3³

= 27

Question 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

1. 4
2. 5
3. 6
4. 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.

WBBSE Class 9 Algebra Solutions Question 7. \(\sqrt[4]{x^{\frac{17}{2}} y^{\frac{5}{2}} \sqrt{x^{\frac{5}{2}} \sqrt{x^{-7} y^6}}}\) = 

1. xy
2. x²y
3. xy²
4. 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

Question 8. \((\sqrt[5]{8})^{\frac{5}{2}} \times(16)^{-\frac{3}{2}}\) = 

1. \(2^{-\frac{9}{2}}\)
2. \(2^{\frac{9}{2}}\)
3. \(2^{-\frac{7}{2}}\)
4. \(2^{\frac{7}{2}}\)

Solution:

\((\sqrt[5]{8})^{\frac{5}{2}} \times(16)^{-\frac{3}{2}}\)

= \(8^{\frac{1}{5} \times \frac{5}{2}^5} \times(16)^{-\frac{3}{2}}\)

= \(\left(2^3\right)^{\frac{1}{2}} \times\left(2^4\right)^{-\frac{3}{2}}=2^{3 \times \frac{1}{2}} \times 2^{4 \times-\frac{3}{2}}=2^{\frac{3}{2}} \times 2^{-6}=2^{\frac{3}{2}-6}=2^{-\frac{9}{2}}\)

Question 9. \(4^{\frac{1}{3}} \times\left[2^{\frac{1}{3}} \times 3^{\frac{1}{2}}\right] \div 9^{\frac{1}{4}}\)

1. 1
2. 2
3. -2
4. 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

Question 10. \(\left\{(125)^{-2} \times(16)^{-\frac{3}{2}}\right\}^{-\frac{1}{6}}\) = 

1. 4
2. 8
3. 10
4. 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

West Bengal Board Class 9 Maths Chapter 6 Laws Of Indices Short Answer Type Questions

Applications of Laws of Indices in Mathematics

Question 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.

Question 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.

Question 3. Find the value of \(\sqrt[3]{\left(\frac{1}{64}\right)^{\frac{1}{2}}}\) 

Solution:

\(\sqrt[3]{\left(\frac{1}{64}\right)^{\frac{1}{2}}}\)

= \(\sqrt[3]{\left(\frac{1}{2^6}\right)^{\frac{1}{2}}}=\sqrt[3]{\left(2^{-6}\right)^{\frac{1}{2}}}=\sqrt[3]{2^{-6 \times \frac{1}{2}}}=\sqrt[3]{2^{-3}}=\left(2^{-3}\right)^{\frac{1}{3}}=2^{-3 \times \frac{1}{3}}=2^{-1}\)

= \(\frac{1}{2}\)

\(\sqrt[3]{\left(\frac{1}{64}\right)^{\frac{1}{2}}}\)  = \(\frac{1}{2}\)

Question 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.

Question 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.

West Bengal Board Class 9 Maths Question 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.

Question 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.

West Bengal Board Class 9 Maths Question 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

Question 9. Simplify: \(\frac{x^{a+b} \cdot x^{a-b} \cdot x^{c-2 a}}{x^{c-a}}\)

Solution:

\(\frac{x^{a+b} \cdot x^{a-b} \cdot x^{c-2 a}}{x^{c-a}}\)

= \(\frac{x^{a+b+a-b+c-2 a}}{x^{c-a}}=\frac{x^c}{x^{c-a}}=x^{c-c+a}=x^a\)

Question 10. Simplify: \(\left\{\left(x^{a+b-c} \times x^{a-b+c}\right)^b\right\}^c\)

Solution:

\(\left\{\left(x^{a+b-c} \times x^{a-b+c}\right)^b\right\}^c\)

= \(\left\{\left(x^{a+b-c+a-b+c}\right)^b\right\}^c=\left\{\left(x^{2 a}\right)^b\right\}^c=\left(x^{2 a b}\right)^c=x^{2 a b c}\)

West Bengal Board Class 9 Maths 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, \(10^{-\frac{1}{4}}, 6^{\frac{1}{3}}, 5^{\frac{1}{2}}\)

WBBSE Class 9 Algebra Solutions 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)

WBBSE Class 9 Algebra Solutions 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)

WBBSE Class 9 Algebra Solutions 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)

WBBSE Class 9 Maths Chapter 6 PDF 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 = \(\frac{4}{7}\)

WBBSE Class 9 Maths Chapter 6 PDF 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.