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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?

  • 3
    Actually, you should have started by raising 5 to both sides of the equation. You know that $a^{\log_a x}=x$ yes?2011-09-21
  • 0
    I don't get what you mean by raising 5. Now I looked the property, still I don't see it being useful here.2011-09-21

1 Answers 1

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

  • 0
    Nice, so that +2 wasn't inside the log there. Thanks.2011-09-21
  • 0
    How do people figure out things like $2=\log_5 25$?2011-09-21
  • 1
    @Kaeser: $2=\log_5 25$ is just another way of saying $5^2=25$.2011-09-21
  • 1
    @Kaeser: If $\log_5 x = 2$ then $5^2 = x$ so $x = 25$.2011-09-21
  • 0
    Right, thanks. That +2 outside the log got me pretty well.2011-09-21