An interesting log problem $3^{{(log_3{x})}^2}+x^{log_3x}=162$

Question

$$3^{{(\log_3{x})}^2}+x^{\log_3x}=162$$

How do I go about doing this. I am stuck at the step $x^{\log_3x} = 81$. Is this right? How do I continue or is it wrong?

Answer 1

This is correct, next step is to take the 3-log of both sides:

$$\log_3(x^{\log_3 x}) = (\log_3 x)(\log_3 x) = \log_3 81 = 4$$

So $\log_3 x = \pm 2$ which means that $x = 3^2 = 9$ or $x=3^{-2} = 1/9$.

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As for the first step that is correct $\log_33^{(\log_3x)^2} = (\log_3x)^2\log_33 = (\log_3x)^2 = \log_3(x^{\log_3 x})$, so the terms on the left hand side are equal.

This means that the first step is complicating things as you could drop the second term instead and get

$$3^{(\log_3x)^2} = 81$$

which seems to lead to a more straight forward solution (but that's only up to a matter of opinion).

Answer 2

$x^{\log_3 x} = 81$. Take $\log_3$ of both sides: $\log_3 x \cdot \log_3 x = \log_3 81 = 4$, so $(\log_3 x)^2 = 4$. You should be able to finish it now.