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


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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.