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?
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
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 $3.8509$.
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 $\log_{10}$:
$$\frac{\log x}{\log 3}\cdot\frac{\log x}{\log 4} \cdot \frac{\log x}{\log 5} = 1$$
$$\left(\log x\right)^3 = \log 3\cdot\log 4\cdot \log 5$$
$$\log x = \left(\log 3\cdot\log 4\cdot \log 5\right)^{1/3}$$
$$x = 10^{\left(\log 3\cdot\log 4\cdot \log 5\right)^{1/3}}$$
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"...
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.
