How to solve $b^x$=$\log_bx$?

Question

I was looking through YouTube when I came across this video by blackpenredpen. The question was$$\text{Solve: }b^x=\log_bx$$which I thought I might be able to solve. Here is my attempt at solving it:$$b^x=\log_bx$$

$$b^{(b^x)}=\require{\cancel}{b^{(\log_bx)}}$$

$$b^{(b^x)}=x$$$$\color{green}{b^x}\ln(b^{\cancel{(b^x)}})=\ln(x)$$$$\color{green}{x}\ln(b\ln(b))=\ln(\ln(x))$$$$\frac{\cancel{x}\ln(b\ln(b))}{\cancel{x}}=\frac{\ln(\ln(x))}{x}$$$$e^{\cancel{\ln}(b\ln(b))}=\sqrt[x]{\ln(x)}$$$$b\ln(b)=e^{\sqrt[x]{\ln(x)}}$$$$\implies\cancel{e}^{\cancel{\ln}(b)}=e^{\frac{\ln(x)}{x}}$$$$b=\sqrt[x]{x}$$$$\therefore x=b^x$$

My question

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Is my solution correct, or what could I do to attain the correct solution more easily?

Answer 1

Note that $y=b^x$ and $y=\log_b x$ are inversed function to each other, so they are mirror symmetric with respect the straight line $y=x$. All intersection points of the two curves $y=b^x$ and $y=\log_b x$ must lie on the straight line $y=x$. So we immediately have

$$b^x=x$$