Find $x$ from logarithmic equation $$ 3^{\log_3^2x} + x^{\log_3x}=162$$
I tried solving this, with basic logarithmic laws, changing base, etc., but with no result, then I went to wolframalpha and it says that its alternate form is: $$2e^{\frac{\log^2x}{\log3}} = 162$$ But I don't know how it came to this result, can you help me guys?
$\endgroup$5 Answers
$\begingroup$The following is how WolframAlpha simplified it: $$\begin{align} 162&=3^{\log_3^2x} + x^{\log_3x} \\&=3^{\log_3x\log_3x} + x^{\log_3x} \\&=\left(3^{\log_3x}\right)^{\log_3x} + x^{\log_3x} \\&=x^{\log_3x} + x^{\log_3x} \\&=2x^{\log_3x} \\&=2\left(e^{\log x}\right)^{\log_3x} \\&=2e^{\log x\log_3x} \\&=2e^{\log x\frac{\log x}{\log 3}} \\&=2e^{\frac{\log^2 x}{\log3}} \end{align}$$
However, this is probably not the optimal process if you want to solve the problem; leaving things in terms of $\log_3x$ is helpful.
$\endgroup$ $\begingroup$$3^{\log^2_3 x} = (3^{\log_3 x})^{\log_3 x} = x^{\log_3 x}$
$\endgroup$ 2 $\begingroup$Since $a^{\log_a M}=M$ we have $x^{\log_3 x}+x^{\log_3 x}=2x^{\log_3 x}=162=2*3^4$.
Hence taking logarithms in base $3$ we have $(\log_3 x)^2=4$ so $(\log_3 x)=2$ (for positives) and $x=9$. Taking negatives we get $x=\frac 19$. There are two solutions.
$\endgroup$ $\begingroup$Introduce a new variable $y$ with $y=\log_3 x$.
Then, we have $3^{y} = x \implies 3^{\log_3^2 x}=3^{y^2},\,\,\text{and}\,\, x^{\log_3 x}=(3^{y})^{y}=3^{y^2}$.
Putting it together we have
$$3^{\log^2_3x}+x^{\log_3 x}=162\implies 2\times 3^{y^2}=162\implies y=\pm 2 \implies x=3^{\pm 2}$$
$\endgroup$ $\begingroup$All those $\log_3$'s make me want to think about what happens when $x$ is a power of $3$. $$ \begin{align} 3^{\log_3^23^k}+\left(3^k\right)^{\log_33^k}&=162\\ 3^{k^2}+3^{k^2}&=162\\ 3^{k^2}&=81\\ k^2&=4\\ k&=\pm2\\ x&=3^2\text{ or }3^{-2} \end{align}$$ (Any positive $x$ can be written as $3^k$, and negative $x$ clearly can't be solutions since the LHS makes no sense for negative $x$.)
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