What is the value of $x$:
\begin{equation} x^{\log_5 x} >5 \end{equation}
Thanks for the help.
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Then take $\log_5$ on both sides and see. – peterwhy Apr 27 '16 at 16:20
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Any value of x greater than 5 – Archis Welankar Apr 27 '16 at 16:26
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@ArchisWelankar: For example: If $x=\frac1{25}$, then $$(\frac{1}{25})^{-2}=25^2>2$$ – Roman83 Apr 27 '16 at 16:42
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$$x^{\log_5x}>5$$ Let $f(x)=x^{\log_5x}-5$ for $x>0$. Find $x$ such that $f(x)=0$ $$\color{red}{\log_5}x^{\log_5x}=\color{red}{\log_5}5$$ $$\log_5x\cdot \log_5x=1$$ $$(\log_5x)^2=1$$ $$\log_5x=\pm1$$ $x=5$ or $x=\frac15$
Answer: $x\in (0;\frac15)\cup(5;+\infty)$
Roman83
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In general, $0\lt a\lt b$ if and only if $\log_5a\lt\log_5b$, and thus
$$\begin{align} 5\lt x^{\log_5x}&\iff\log_55\lt\log_5(x^{\log_5x})\\ &\iff1\lt(\log_5x)^2\\ &\iff1\lt\log_5x\quad\text{or}\quad\log_5x\lt-1\\ &\iff5\lt x\quad\text{or}\quad 0\lt x\lt5^{-1} \end{align}$$
Remark: This is essentially the same solution as in Roman83's answer, just presented in a style that makes the logical connections explcit.
Barry Cipra
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