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Why is it that the rule $$ \log_b(x) \ = \ \frac{\log_c(x)}{\log_c(b)} \ $$ (the logarithm base change rule) is true but $$ \ a \log_b(x) \ = \ \frac{a \log_c(x) }{ a \log_c(b) } \ $$ isn't?

For example why does the equation, $$ \log_{49} 3 \ = \ \frac{\log 49 }{ \log 3 \ } $$ work but $$ \ 4 \log_{49} 3 \ = \ \frac{4 \log 49 }{ 4 \log 3} $$ does not?

All you are doing is multiplying the logarithm by a term, and still multiplying each part by it when "splitting the logarithm up."

P.S. If the answer could be explained in somewhat simple terms it would be appreciated. I'm not like a math major or anything. :)

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    $\begingroup$ You are multiplying by $ \ 4 \ $ on the left-hand side of the equation, but by $ \ \frac{4}{4} \ $ on the right-hand side. That might be why this doesn't work... $\endgroup$ – colormegone Dec 23 '15 at 1:08
  • $\begingroup$ Please check whether I've edited your expressions correctly. If I interpreted what you typed properly, the last two expressions are not equations for other reasons... $ \ \log 49^3 \ $ in any base is $ \ 3 \log 49 \ $ , rather than what you have. $\endgroup$ – colormegone Dec 23 '15 at 1:20
  • $\begingroup$ @RecklessReckoner Well they were supposed to be "four times log base 49 of three" as in the case of the last one (like Dr. Sonnhard wrote his), not "log base 49 raised to the third power." Ah, but I see now what you mean about multiplying by 4/4 on the right side. I bet you why that's it! $\endgroup$ – Guy Dec 23 '15 at 1:40
  • $\begingroup$ OK, but I just carried over the " ^ " that was already in those lines. I'll revise that now. $\endgroup$ – colormegone Dec 23 '15 at 1:44
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You have $$\log_b(x) = \frac{\log_c(x)}{\log_c(b)}.$$

Multiplying on both sides by $a$ gives

$$\begin{align} a\log_b(x) &= a\left(\frac{\log_c(x)}{\log_c(b)}\right)\\ &= \frac a1\left(\frac{\log_c(x)}{\log_c(b)}\right)\\ &= \frac{a\log_c(x)}{\log_c(b)}. \end{align}$$

Your mistake was to multiply on the left by $a$ but on the right by $\frac aa$, which is equal to $1$.

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  • $\begingroup$ Ah, okay thank you Dr. Sonnhard's answer was first (and I think he was saying the same thing) but yours was clearer and explained well. I understand now and I checked the math with removing the coefficient 4 from the denominator and it worked. Thanks! $\endgroup$ – Guy Dec 23 '15 at 1:52
  • $\begingroup$ You're welcome! Glad to help. $\endgroup$ – Théophile Dec 23 '15 at 1:53
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you can only write $a\log_bx=\frac {a\log_cx}{\log_c b}$ for $a\ne 0$

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  • $\begingroup$ Why is 'a' a coefficient of the numerator logarithm but not of the denominator? $\endgroup$ – Guy Dec 23 '15 at 1:36
  • $\begingroup$ since it is the equation not true $$\frac{a\log_c x}{a\log _c b}=\frac{\log_c x}{\log_c b}$$ if $a \ne 0$ $\endgroup$ – Dr. Sonnhard Graubner Dec 23 '15 at 1:38
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Let me explain it using the logarithmic rules:

Since $$ \log_b(x) = \frac{\log_c(x)}{\log_c(b)}\qquad\text{ and }\qquad a\log_b(x^a)$$ we have $$ a\log_b(x) =\log_b(x^a) = \frac{\log_c(x^a)}{\log_c(b)}=\frac{a\log_c(x)}{\log_c(b)} $$ So, $a$ is a coefficient only of the numerator.

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