Solve for $x$: question on logarithms. The question:
$$\log_3 x \cdot \log_4 x \cdot \log_5 x = \log_3 x \cdot \log_4 x \cdot \log_5 x \cdot \log_5 x \cdot \log_4 x \cdot \log_3 x$$
My mother who's a math teacher was asked this by one of her students, and she can't quite figure it out. Anyone got any ideas?
 A: Use the identity 
$$
\log_a x=\ln x/\ln a.
$$
A: Following up on Jaeyong Chung's answer, and working it out:
$$ 1 =\log_3x\log_4x\log_5x$$
$$1=\frac{(\ln x)^3}{\ln3\ln4\ln5}$$
$$(\ln x)^3 = \ln3\ln4\ln5$$
$$(\ln x) = \sqrt[3]{\ln3\ln4\ln5}$$
$$x = \exp\left(\sqrt[3]{\ln3\ln4\ln5}\right) \approx 3.85093$$
EDIT: And, of course, the obvious answer that everyone will overlook: $x=1$ makes both sides of the equation zero. :D
A: Given:
$$\log_3 x \cdot \log_4 x \cdot \log_5 x = \log_3 x \cdot \log_4 x \cdot \log_5 x \cdot \log_5 x \cdot \log_4 x \cdot \log_3 x$$
One immediately obvious solution is $x = 1$.  Regardless of the base $b$, $\log_b 1 = 0$. So $x = 1$ is a solution: it nullifies all factors simultaneously, making the equation true.
So on to chasing other solutions.
Firstly, note the repeating factors in right side, which condense to a square term:
$$\log_3 x \cdot \log_4 x \cdot \log_5 x = \left(\log_3 x \cdot \log_4 x \cdot \log_5\right)^2$$
Substitute z for the repeated subexpression: let $z = \log_3 x \cdot \log_4 x \cdot \log_5 x$. We then get a simplified form which clarifies the relationship:
$$z = z^2$$
This quadratic has two solutions:
$$z \in \lbrace 0, 1 \rbrace$$
Which corresponds to these two cases when we substitute back the original log factors for $z$:
$$\log_3 x \cdot \log_4 x \cdot \log_5 x \in \lbrace 0, 1 \rbrace$$
But the zero case corresponds to the $x = 1$ solution we already know, so henceforth we only care about the second case:
$$\log_3 x \cdot \log_4 x \cdot \log_5 x = 1$$
We can convert the logs to a common base, arbitrarily picking 3:
$$\log_3 x \cdot \frac{\log_3 x}{\log_3 4} \cdot \frac{\log_3 x}{\log_3 5} = 1$$
$$\frac{\left(\log_3x\right)^3}{\log_3 4\cdot \log_3 5} = 1$$
$$\left(\log_3x\right)^3 = {\log_3 4\cdot \log_3 5}$$
$$\log_3x = \left(\log_3 4\cdot \log_3 5\right)^{1/3}$$
$$x = 3^{\left(\log_3 4\cdot \log_3 5\right)^{1/3}}$$
This is approximately $3.8509$.

Appendix:
If the aim is to get a decimal figure with a calculator, it's better to use base 10 as the common base rather than 3, and this base is also better than $e$.  We can then use a calculator which provides only a base 10 log function, and an $x^y$ button, which are more common than support for natural log, and a base $e$ exp function, or the availability of $e$ as a constant. Below, it is to be understood that $\log$ refers to $\log_{10}$:
$$\frac{\log x}{\log 3}\cdot\frac{\log x}{\log 4} \cdot \frac{\log x}{\log 5} = 1$$
$$\left(\log x\right)^3 = \log 3\cdot\log 4\cdot \log 5$$
$$\log x = \left(\log 3\cdot\log 4\cdot \log 5\right)^{1/3}$$
$$x = 10^{\left(\log 3\cdot\log 4\cdot \log 5\right)^{1/3}}$$
A: The "change-of-base" formula should at least allow you to reduce this to
$$\log_3 x \cdot \left(\frac{\log_4 x}{\log_4 3}\right) \cdot \left(\frac{\log_5 x}{\log_5 3}\right) \ =  \ (\log_3 x)^3 \ = \ \frac{1}{\log_4 3 \cdot \log_5 3} . $$
The number won't be "pretty"...
A: As above, use $$log_a x=\frac{\ln x}{\ln a}$$
Hence, $$\log_3 x \cdot \log_4 x \cdot \log_5 x  =  1$$ would become $$\frac{\ln x \cdot \ln x \cdot \ln x}{\ln 3 \cdot \ln 4 \cdot \ln5}=1$$
Thus, $$(\ln x)^3=\ln 3 \cdot \ln 4 \cdot \ln 5=2.45117$$
Hence, $$\ln x = 1.34831$$
This gives $x$ as $$x=e^{1.34831}=3.85091$$ approximately.
EDIT: Here, we assume that $x \neq 1$. So $x=1$ is also a possible solution.
