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find $x$ : $${\log_3(\log_2 x)}=\log_2(\log_3x)$$

My try :

$$f(x):=\log_3 x \ \ , \ \ g(x):=\log_2x$$

$$g∘f: \{x\in D_f :f(x)\in D_g\}\to \mathbb{R}\\ g(f(x))=\log_2(\log_3x)$$

and :

$$f∘g: \{x\in D_g:f(x)\in D_f\}\to \mathbb{R}\\f(g(x))=\log_3(\log_2x)$$

now ?

Almot1960
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3 Answers3

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we set $$a=\log_2 x$$ and $$b=\log_3 x$$ and our equation is now $$\frac{\ln(a)}{\ln(3)}=\frac{\ln(b)}{\ln(2)}$$ now we have $$\ln(2)\ln(a)=\ln(3)\ln(b)$$ thus $$\ln(2)\ln(\log_2 x)=\ln(3)\ln(\log_3 x)$$ or $$ \ln(2)\ln\left(\frac{\ln(x)}{\ln(2)}\right)=\ln(3)\ln\left(\frac{\ln(x)}{\ln(3)}\right)$$ now Setting $$\ln(x)=t$$ and we get $$\ln(2)\left(\ln(t)-\ln(\ln(2))\right)=\ln(3)\left(\ln(t)-\ln(\ln(3))\right)$$ this can be solved fir $\ln(t)$ etc

can you finish this?

Dr. Sonnhard Graubner
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Using Logarithm change of base formula ... $\log_b(y)=\frac{\ln(y)}{\ln(b)}$ ... twice \begin{eqnarray*} \frac{\ln \left(\frac{\ln(x)}{\ln(3)}\right)}{\ln(2)} = \frac{\ln \left(\frac{\ln(x)}{\ln(2)}\right)}{\ln(3)} \end{eqnarray*}

After a some algebra \begin{eqnarray*} \ln(\ln(x)) = \frac{\ln(\ln(3)) \times \ln(3) -\ln(\ln(2)) \times \ln(2)}{\ln(3)-\ln(2)} \end{eqnarray*} We get $x=\color{red}{11.18 \cdots}$.

Donald Splutterwit
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$$\log_3 \log_2 x = \log_2 \log_3 x$$

First let's convert everything to natural logs, using the change of base formula: $\log_b a = \dfrac{\ln a}{\ln b}$

$$ \log_3 \log_2 x = \frac{\ln (\log_2 x)}{\ln 3} = \frac1{\ln 3}\ln\left(\frac{\ln x}{\ln 2}\right) = \frac1{\ln 3}(\ln\ln x) - \ln\ln 2) $$ Note that I also used the log rule that says $\ln(A/B) = \ln A - \ln B$. Similarly, we have $$ \log_2 \log_3 x = \frac1{\ln 2}(\ln\ln x - \ln\ln 3) $$

So then \begin{align*} \log_3 \log_2 x &= \log_2 \log_3 x\\[0.3cm] \frac1{\ln 3}(\ln\ln x - \ln\ln 2) &= \frac1{\ln 2}(\ln\ln x - \ln\ln 3)\\[0.3cm] \ln 2 (\ln\ln x - \ln\ln 2) &= \ln 3 (\ln\ln x - \ln\ln 3)\\[0.3cm] \ln 2 \cdot \ln\ln x - \ln 2 \cdot \ln\ln 2 &= \ln 3 \cdot \ln\ln x - \ln 3 \cdot \ln\ln 3\\[0.3cm] \ln 2 \ln\ln x - \ln 3 \ln\ln x &= \ln 2 \ln\ln 2 - \ln 3 \ln\ln 3\\[0.3cm] (\ln 2 - \ln 3) \ln\ln x &= \ln 2 \ln\ln 2 - \ln 3 \ln\ln 3\\[0.3cm] \ln\ln x &= \frac{\ln 2 \ln\ln 2 - \ln 3 \ln\ln 3}{\ln 2 - \ln 3}\\[0.3cm] \ln x &= \exp\left(\frac{\ln 2 \ln\ln 2 - \ln 3 \ln\ln 3}{\ln 2 - \ln 3}\right)\\[0.3cm] x &= \exp\left[\exp\left(\frac{\ln 2 \ln\ln 2 - \ln 3 \ln\ln 3}{\ln 2 - \ln 3}\right)\right] \end{align*}