How to solve this? My problem is to solve:
$$x^{\log_3(x-4)} \ge \frac{1}{27}.$$
The log base is $3$.
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3Is that exponent $(\log x) - 4$ or $\log(x - 4)$? – Jack M Mar 01 '14 at 20:56
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2If it is $\log(x) - 4$, then you should get a quadratic: $\log_{3}(x)(\log_{3}(x)-4) \geq -3$, since $\log$ is strictly increasing. – Christopher K Mar 01 '14 at 20:57
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log(x-4) and base is 3 – user2511140 Mar 01 '14 at 20:57
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Take $\log_3$ from both sides. – Lucian Mar 01 '14 at 20:58
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@Lucian This user wants to determine $x^{\log(x)}$ – NasuSama Mar 01 '14 at 20:58
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@NasuSama The original question said "my problem in solve $x^{\log x}$". Given that he then states an inequality for $x^{\log x}$ which happens to have a very clean solution for $x$ (see the answer below), I think it is fair to assume he just wanted to solve for $x$. – TMM Mar 02 '14 at 00:20
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Following the suggestions above, we take the log base 3 of both sides to get $(\log_{3}x)(\log_{3}x-4)\ge-3$, using the fact that $f(x)=\log_{3}x$ is increasing. Substituting $u=\log_{3}x$ gives $u(u-4)\ge-3$ so $u^2-4u+3\ge0$.
Then $(u-3)(u-1)\ge0$, so either $u\le1$ or $u\ge3$.
If $u\le1$, then $\log_{3}x\le1$ so $0<x\le3$.
If $u\ge3$, then $\log_{3}x\ge3$ so $x\ge27$.
Therefore the solution is given by $(0,3]\cup[27,\infty)$.
user84413
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