<s>[INST] Exponential algebra problem We need to solve for x:
$$54\cdot 2^{2x}=72^x\cdot\sqrt{0.5}$$
My proposed solution is below.
 [/INST] Let's do this in another way:
$$3^3\cdot 2^{2x+1}=2^{3x-0.5}\cdot 3^{2x}$$
$$2^{-x+1.5}=3^{2x-3}$$
Let's take a log with base 2:
$$\log_{2}{2^{-x+1.5}}=\log_{2}{3^{2x-3}}$$
$${-x+1.5}=\log_{2}{3^{2x-3}}$$
Let's move to base 3:
$$-x+1.5=\frac{\log_3{3^{2x-3}}}{\log_3{2}}$$
$$-x+1.5=\frac{2x-3}{\log_3{2}}$$
$$-x(1+\frac{2}{\log_32})=-1.5-\frac{3}{\log_32}$$
$$x=1.5$$
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