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

Updated On: Sep 13, 2024
  • -20
  • -12
  • -24
  • -16
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The Correct Option is B

Solution and Explanation

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

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

\(⇒2^{(\frac{19}{2}+4+3n)}\times3^{(4+2m)}=2^{(4m+2)}\times3^n\)
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.

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