Taken from the PAT (Physics Aptitude Test):
$\log_x 25 = \log_5 x$
I've tried making them the same base but I ended up with:
$\log_5 x (\log_5 x)= 2$
or
$\log_x 25(\log_x 5) = 1$
which I then simplified to
$\log_x 5 (\log_x 5 + 1) = 1$
Not sure if the last line is correct math but I'd appreciate some help on how to solve this equation.
On
A long way:
We want to find $x$ such that: $$\log_x 25 = \log_5 x$$
I would convert both logs to the same base, and choose the common base to be $10$ in an attempt to simplify calculations. Here I write $log(x)$ to mean $log_{10}(x)$ again to simplify the looks.
$$\log_x 25 = \frac{log(25)}{log(x)}$$
$$\log_5 x = \frac{log(x)}{log(5)}$$
Since the L.H.S is equal in both of the above equations, the R.H.S is equal, we get:
$$\frac{log(25)}{log(x)}=\frac{log(x)}{log(5)}$$
Hence (we have to assume $x is not 1 because log(1)=0):
$$log(25)log(5) =log(x)log(x)$$
$$log(x)=\sqrt{log(5)log(25)}=0.98849285982$$
Now we raise both sides to power of $10$ since we use $log$ to the base $10$:
$$x=9.73851774234$$
To verify:
$$\log_{9.73851774234} 25 = \log_{5} (9.73851774234)$$
$$ 1.4142135624 = 1.4142135624$$
In fact there is another value of $x$ that satisfies the equation:
$$x=0.103$$
I obtained this value graphically. You can see: Desmos Plot.
$\log_x 25 = 2\log_x 5=\frac{2}{\log_5 x}$ so $(\log_5 x)^2=2$, $\log_5 x=\pm \sqrt{2}$, $x=5^{\sqrt{2}}$, $x=5^{-\sqrt{2}}$
Verification: $\log_5 5^{\sqrt{2}}=\sqrt{2}$; $\log_{5^{\sqrt{2}}}25=\frac{1}{\sqrt{2}}\log_5 25=\frac{2}{\sqrt{2}}=\sqrt{2}$
$\log_5 5^{-\sqrt{2}}=-\sqrt{2}$; $\log_{5^{-\sqrt{2}}}25=-\frac{1}{\sqrt{2}}\log_5 25=-\frac{2}{\sqrt{2}}=-\sqrt{2}$