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How can I find the inverse of $3^{5^{x}}$ ? I tried using logarithm in base 3: $3^{5^{y}}=x \Longrightarrow \log_3x=5^y \Longrightarrow \log_5(\log_3x)=y$? Is it correct? in my book it says its another answer from those given so I can't know the correct one.Answers in my book are:

a)$\log_{243}x \quad x \in(0,\infty)$

b)$\log_{15}x$

c)$\log_{243}x \quad x \in(1,\infty)$

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  • $\begingroup$ What does the book say the answer is? Or does it just say "none of the above"? $\endgroup$ – tilper Jul 8 '17 at 14:59
  • $\begingroup$ none of the answers above $\endgroup$ – Lola Jul 8 '17 at 14:59
  • $\begingroup$ What are the other choices? $\endgroup$ – tilper Jul 8 '17 at 15:03
  • $\begingroup$ I edited the question $\endgroup$ – Lola Jul 8 '17 at 15:09
  • $\begingroup$ See the side note in my answer. If the problem really says $3^{5^x}$ and not $(3^5)^x$ then there's some unfortunate ambiguity here. Probably best to clarify with your instructor or at least see if your book has an explanation somewhere. $\endgroup$ – tilper Jul 8 '17 at 15:19
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Step 1: Write $y = 3^{5^x}$.

Step 2: Swap $x$ and $y$ to get $x = 3^{5^y}$.

Step 3: Solve for $y$. \begin{align*} x &= 3^{5^y}\\ \log_3 x &= \log_3 \left(3^{5^y}\right)\\ \log_3 x &= 5^y\\ \log_5 (\log_3 x) &= \log_5 (5^y)\\ \log_5 (\log_3 x) &= y \end{align*} So it looks like the answer is indeed $y = \log_5(\log_3 x)$.

Note that as another answerer pointed out, one of the choices may be this answer in a different form.


Side note: Some people may argue that the original function can be interpreted as $y = 3^{5^x} = (3^5)^x = 243^x$. I disagree with this because I interpret $3^{5^x}$ without parentheses as meaning $3^{(5^x)}$, which I believe more closely conforms to order of operations.

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  • $\begingroup$ Though you disagree, this is clearly what the problem statement means. Indeed the three proposals are of the form $\log_nx$, so that the original function is a simple exponential $n^x$. I'd be curious to see the typesetting in the book. $\endgroup$ – Yves Daoust Jul 8 '17 at 15:41
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Well, there are multiple possible ways to express the answer to this problem. If $$y=3^{5^x}$$ then you could solve for $x$ your way, by using $\log_3$ and $\log_5$ to get $$x=\log_5 \log_3 x$$ or you could use the natural logarithm (which is what I suspect your book did): $$y=3^{5^x}$$ $$\ln y=\ln 3^{5^x}$$ $$\ln y=5^x\ln 3$$ $$\ln \ln y=\ln(5^x\ln 3)$$ $$\ln \ln y=\ln5^x+\ln \ln 3$$ $$\ln \ln y-\ln \ln 3=x\ln5$$ $$x=\frac{\ln \ln y-\ln \ln 3}{\ln 5}$$ Is this close to what your book had?

The book may also have chosen to deviate from this process at this step: $$\ln y=5^x\ln 3$$ and instead divide both sides by $\ln 3$ to get $$\frac{\ln y}{\ln 3}=5^x$$ $$\ln \frac{\ln y}{\ln 3}=\ln 5^x$$ $$\ln \frac{\ln y}{\ln 3}=x\ln 5$$ $$x=\frac{1}{\ln 5}\ln \frac{\ln y}{\ln 3}=$$ Again, all of these answers are equivalent, you yours is still correct.

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