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1. $2^{3.6} \times 4^{3.6} \times 4^{3.6} \times (32)^{2.3} =$ ${(32)}^{?}$

A. 5.9

B. 7.7

C. 9.5

D. 13.1

Solution:

\begin{aligned} Let 2^{3.6} \times 4^{3.6} \times 4^{3.6} \times (32)^{2.3} = {(32)}^{x} \\ Then, 2^{3.6} \times {(2^{2})}^{3.6} \times {(2^{2})}^{3.6} \times (2^{5})^{2.3} = {(2^{5})}^{x} \\ \Leftrightarrow 2^{3.6} \times 2^{(2 \times 3.6)} \times 2^{(2 \times 3.6)} \times (2^{5})^{2.3} = {(2^{5})}^{x} \\ \Leftrightarrow 2^{(3.6 + 7.2 + 7.2)} \times (2^{5})^{2.3} = {(2^{5})}^{x} \\ \Leftrightarrow 2^{18} \times (2^{5})^{2.3} = {(2^{5})}^{x} \\ \Leftrightarrow {(2^{5})}^{3.6} \times (2^{5})^{2.3} = {(2^{5})}^{x} \\ \Leftrightarrow {(2^{5})}^{(3.6 + 2.3)} = {(2^{5})}^{x} \\ \Leftrightarrow {(2^{5})}^{5.9} = {(2^{5})}^{x} \\ \Leftrightarrow x = 5.9 \end{aligned}

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