Question 26.

We have, $(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$

Converting both sides in powers of 2 and 3, we get

$2^{\frac{19}{2}}3^42^43^{2m}2^{3n}$ = $3^n2^{4m}2^{\frac{6}{4}}$

Comparing the power of 2 we get, $\frac{19}{2}+4+3n=4m+\frac{6}{4}$

=> 4m=3n+12 .....(1)

Comparing the power of 3 we get, $4+2m=n$

Substituting the value of n in (1), we get

4m=3(4+2m)+12

=> m=-12