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log base(16) 32 = x + 2.
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\[\LARGE \log_{16}(32)=x+2\] \[\LARGE \log_{2^4}(2^5)=x+2\] \[\LARGE \frac14 \cdot 5 \cdot \log_{2}(2)=x+2\] \[\LARGE \frac54 \cdot 1=x+2\] can you do it now ? :)
raise both sides to powers of 16 32 = 16^(x+2) 32 = 16^(5/4) then 16^(5/4) = 16^(x + 2) equate the powers 5/4 = x + 2 x = -3/4
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