# Find the Values of M and N If : 4^(2m) = ( Root(3)(16))^(-6/N) = (Sqrt8)^2 - Mathematics

Sum

Find the values of m and n if :
4^(2m) = ( root(3)(16))^(-6/n) = (sqrt8)^2

#### Solution

4^(2m) = ( root(3)(16))^(-6/n) = (sqrt8)^2
⇒ 4^(2m) = (sqrt8)^2                    ....(1)

and
(root(3)(16))^(-6/n) = (sqrt8)^2     ....(2)
From (1)
4^(2m) = (sqrt8)^2

⇒ (2^2)^(2m) = (sqrt(2^3))^2

⇒ 2^(4m) = [(2^3)^(1/2)]^2

⇒ 2^(4m) = [ 2^( 3 xx 1/2 )]^2

⇒ 2^(4m) =  2^( 3 xx 1/2 xx 2)

⇒ 2^(4m) = 2^3

⇒ 4m = 3

⇒ m = 3/4

From (2), We have
(3sqrt(16))^(-6/x) = (sqrt8)^2

⇒ ( root(3)(2 xx 2 xx 2 xx 2))^(-6/x) = (sqrt( 2 xx 2 xx 2))^2

⇒ ( root(3)(2^4))^(-6/x) = ( sqrt(2^3))^2

⇒ [(2^4)^(1/3)]^(-6/x) = [(2^3)^(1/2)]^2

⇒ [2^(4/3)]^(-6/x) = [2^(3/2)]^2

⇒ 2^( 4/3 xx ( - 6/x ) = 2^(3/2 xx 2)

⇒ 2^(-8/x) = 2^3

⇒ -8/x = 3

⇒  x = -8/3 "Thus m" = 3/4x = - 8/3

Concept: Solving Exponential Equations
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#### APPEARS IN

Selina Concise Mathematics Class 9 ICSE
Chapter 7 Indices (Exponents)
Exercise 7 (B) | Q 5 | Page 100