Question

(3/4)n-¹ × (16/9)¹-n = 27/64. find the value of n​

Answer 1

Answer: 7 value of n

Step-by-step explanation:

Answer 2

HEY,

The question seems like $\bold{(3/4)^{n-1} \times (16/9)^{1-n} = 27/64}$

$\bold{\rightarrow (3/4)^{n-1} \times (4/3)^{2(1-n)} = 27/64}$

We know m^n = 1/m^-n and x^n*x^m = x^(n+m)

so, $\bold{\rightarrow (3/4)^{n-1} \times (3/4)^{-2+2n} = 27/64}$

[ (x/y)^n = (y/x)^(-n)]

$\bold{\rightarrow (3/4)^{n-1-2+2n} = (3/4)^3}$

$\bold{\rightarrow (3/4)^{3n-3} = (3/4)^3}$

we know if x^m = x^n , then m = n

$\bold{\therefore , 3n-3 = 3}$

$\bold{3n = 6}$

$\bold{ n = \frac{6}{3}}$

$\bold{n = 2}$

Verification,

put n = 2 ,

then, L.H.S = (3/4)^(2-1) × (16/9)^(1-2)

$\rightarrow$ (3/4) × (16/9)^-1

$\rightarrow$ (3/4) × (9/16)

$\rightarrow$ 27/64 = R.H.S