If $x$ is a positive quantity such that $2^x = 3^{\log_5 2}$, then $x$ is equal to

Given: $2^x = 3^{\log_5 2}$ where $x > 0$

Find: The value of $x$

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When we have exponential equations like this, taking logs is usually the fastest approach!

Taking $\log_5$ of both sides:

$\log_5(2^x) = \log_5(3^{\log_5 2})$

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Using the property $\log_a(b^c) = c \cdot \log_a(b)$:

Left side: $\log_5(2^x) = x \cdot \log_5(2)$

Right side: $\log_5(3^{\log_5 2}) = (\log_5 2) \cdot \log_5(3)$

So our equation becomes:

$x \cdot \log_5(2) = (\log_5 2) \cdot \log_5(3)$

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Since $\log_5(2) \neq 0$ (because $2 \neq 1$), we can divide both sides by $\log_5(2)$:

$x = \dfrac{(\log_5 2) \cdot \log_5(3)}{\log_5(2)} = \log_5(3)$

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Using the property that $\log_5(3) = \log_5(3) - \log_5(5) + \log_5(5)$:

$\log_5(3) = \log_5\left(\dfrac{3}{5}\right) + 1$

Therefore:

$x = 1 + \log_5\left(\dfrac{3}{5}\right)$