Proof of logarithmic identity $\log_g x=\log_a x\cdot\log_g a$

Question

I have to prove the alleged link between the logarithms in base g and a

$$\log_g x=\log_a x\cdot\log_g a$$

I know that this can be written as:

$$\frac{\ln x}{\ln g}=\frac{\ln x}{\ln g}$$

But does this prove the alleged link or do i have to do something further?

I know that this is derived from: $$\log_g x=\frac{\ln x}{\ln g}$$

Answer 1

Using your last formula $\left(\log_g(x)=\frac{\ln(x)}{\ln(g)}\right)$

$$\frac{\log_g(x)}{\log_a(x)}=\frac{\frac{\ln(x)}{\ln(g)}}{\frac{\ln(x)}{\ln(a)}}=\frac{\ln(x)\cdot \ln(a)}{\ln(x)\cdot \ln(g)}=\frac{\ln(a)}{\ln(g)}=\log_g(a)\iff \log_g(x)=\log_g(a)\log_a(x).$$

Q.E.D.

Answer 2

$$g^{\log_{a}x\times\log_{g}a}=\left(g^{\log_{g}a}\right)^{\log_{a}x}=a^{\log_{a}x}=x=g^{\log_{g}x}$$ hence: $$\log_{a}x\times\log_{g}a=\log_{g}x$$