If m and n are integers such that (\sqrt2)^19 3^4 4^2 9^m 8^n=3^n 16^m(\sqrt^4 {64}) then m is

CAT 2019 Shift 1 Paper

Question : 79 of 100

Marks: +1, -0

If m and n are integers such that

(√2)^19 3^4 4^2 9^m 8^n=3^n 16^m(√^4 {64})

then m is

Solution:

(√{2})^{19} 3^{4} 4^{2} 9^{m} 8^{n}=3^{n} 16^{m}(√^{4}{64})

⇒ 2^{19{/} 2} × 3^{4} × 2^{4} × 3^{2 m} × 2^{3 n}=3^{n} × 2^{4 m} × 2^{3 {/} 2}

⇒ 2^{(19 {/} 2+4+3 n)} × 3^{(4+2 m)}=2^{(4 m+3)} × 3^{n}

Comparing the powers of same bases we get

\;{19}/{2}+4+3 n=4 m+\;{3}/{2} ⋅s

4+2 m=n ⋅s(2)

Substitute the value of

n

from (2) in (1) and solving for

m,

we get

m=-12

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