Find the integer root of the logarithmic equation: $\log_2\left(\frac x5\right)+\log_x\left(\frac 25\right)=\log 4$ This is a contest question asked 6-7 months ago in a real-time mathematics competition with high-school students. Unfortunately, since the source is not in English, I cannot add it here.

Find the integer root of the logarithmic equation:
$$\log_2\left(\frac x5\right)+\log_x\left(\frac 25\right)=\log 4$$
where, $\log (x)=\log_{10} (x),x>0.$

I have attempts to solve the question
$$\log_2x-\log_25+\log_x2-\log_x 5=\log4$$
I know that,
$$\log_x2=\frac{1}{\log_2x}\\
\log_x5=\frac{1}{\log_5x}\\$$
Based on the logarithm rules above, we obtain
$$\log_2x-\log_25+\log_x2-\log_x 5=\log4\\
\log_2x+\frac{1}{\log_2x}-\frac{1}{\log_5x}=\log4+\log_25$$
But, I can't think of anything here. My goal was to make the base of the logarithm not dependent on the variable.  But that didn't work.
Do you have any solution suggestions?
 A: Let, $x=10^{u}, \thinspace u \neq 0$ then we have
$$
\begin{align}&\frac{u}{\log 2}-\frac{\log 5}{\log 2}+\frac{1}{u}(\log 2-\log 5)=\log 4 \\
\implies &u^{2}-u \log 5+\log 2\left(\log 2-\log 5\right)=u \log 2 \log 4 \\
\implies &u^{2}-u\left(\log 5+2 \log ^{2} 2\right)+\log 2\left(\log 2-\log 5\right)=0 \\
\implies &u^{2}-u\left(2 \log ^{2} 2-\log 2+1\right)+\left(2 \log ^{2} 2-\log 2\right)=0 \end{align}
$$
Then let $m=2 \log ^{2} 2-\log 2$, we get
$$
\begin{align}&u^{2}-u(m+1)+m=0 \\ 
\implies &\Delta=(m+1)^{2}-4 m=(m-1)^{2} \\
\implies &u_{1,2}=\frac{m+1 \pm(m-1)}{2}\end{align}
$$
Finally, we conclude that
$$
\begin{align}
&u_1=\frac{m+1-m+1}{2}=1\\
\implies &x_1=10^{1}=10\end{align}
$$
and
$$
\begin{align}&u_2=\frac{m+1+m-1}{2}=m\\
\implies &x_2=10^{m}=10^{2 \log ^{2} 2-\log 2}.\end{align}
$$

Small Supplement:
Our equation has $1$ rational and
$1$ irrational root. The rational root $x=10$ is also an integer root.
The other root is the irrational root and can be expressed as follows:
$$x=10^{2 \log ^{2} 2-\log 2}.$$
A: Hint : Choose a common base for all the logarithms (the natural one for instance). Then rewrite using $\log_a(b) =\frac{\ln(a)}{\ln(b)}$ and $\ln(a/b) = \ln(a)-\ln(b)$. This gives you a quadratic equation in $X =\ln(x)$, which you can easily solve.
A: Use the fact that, $\log_{a}(b)=\frac{\ln a}{\ln b}$
The equation can be written as,
$$\frac{\ln(x/5)}{\ln2}+\frac{\ln(2/5)}{\ln x}$$
Use the quotient rule for logarithms and separate $\ln(x)$ and substitute $u=\ln(x)$ we get a quadratic equation. Solving for $u$ gives the answer.
