# 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?

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Actually, you should have started by raising 5 to both sides of the equation. You know that $a^{\log_a x}=x$ yes? –  Ｊ. Ｍ. Sep 21 '11 at 13:12
I don't get what you mean by raising 5. Now I looked the property, still I don't see it being useful here. –  Kaeser Sep 21 '11 at 13:16

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

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Nice, so that +2 wasn't inside the log there. Thanks. –  Kaeser Sep 21 '11 at 13:40
How do people figure out things like $2=\log_5 25$? –  Kaeser Sep 21 '11 at 13:41
@Kaeser: $2=\log_5 25$ is just another way of saying $5^2=25$. –  Ｊ. Ｍ. Sep 21 '11 at 13:43
@Kaeser: If $\log_5 x = 2$ then $5^2 = x$ so $x = 25$. –  TMM Sep 21 '11 at 13:44
Right, thanks. That +2 outside the log got me pretty well. –  Kaeser Sep 21 '11 at 13:47