Why can't you add a coefficient before the logarithm base change rule?

Question

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. :)

Answer 1

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 $1$.

Answer 2

you can only write $a\log_bx=\frac {a\log_cx}{\log_c b}$ for $a\ne 0$

Answer 3

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.