Question

If m and n are integers such that $(\sqrt{2})^{19} 3^{4} 4^{2} 9^{m} 8^{n}= 3^{n} 16^{m} (^4\sqrt{64})$ then m is

• A. -20
• B. -12
• C. -24
• D. -16

Answer 1

Answer: (B)

-12

Answer 2

$(\sqrt{2})^{19}3^44^29^m8^n=3^n16^m(\sqrt[4]{64})$

$⇒2^{\frac{19}{2}}\times3^4\times2^4\times3^{2m}\times2^{3n}=3^n\times2^{4m}\times2^{\frac{2}{3}}$

$⇒2^{(\frac{19}{2}+4+3n)}\times3^{(4+2m)}=2^{(4m+2)}\times3^n$
By comparing the exponents of identical bases, we obtain
$\frac{19}{2}+4+3n=4m+\frac{3}{2}.....(1)$

$4+2m=n.....(2)$
Replace the value of n from equation (2) into equation (1) and, upon solving for m, we obtain m = -12.