Find M, since $\log_5 M = 2\log_5 A - \log_5 B+2$ Find M, since $\log_5 M = 2\log_5 A - \log_5 B+2$
I tried this:
The answer is in function of A and B.
$\frac{\log_M M}{\log_M 5} = 2\frac{\log_M A}{\log_M 5} - \frac{\log_M B+2}{\log_M 5}$
$1=2\log_M A - \log_M B+2$
$\log_M A^2 = \log_M B+2$
$A^2=B+2$
$\log_5 M = 2\log_5 A - \log_5 A^2$
$\log_5 M = 2\log_5 A - 2 \log_5 A$
$\log_5 M = 0$
$5^0 = M \implies M=1 $
So I don't find how to get an answer in function of A and B nor there is 1 as answer. What did I do wrong?

$$\log_5 M = 2 \log_5 A − \log_5 B + 2 = \log_5 A^2 + \log_5 \frac{1}{B} + \log_5 25 = \log_5 \left(\frac{25 A^2}{B}\right) \quad \Longrightarrow \quad M = \frac{25A^2}{B}$$
