# Logarithmic property equivalent to $b^x=y \Rightarrow \log_{b}(y)=x$

I was trying to solve $(1.05)^t=100$.

So I used the logarithmic rule I know: if $b^x=y \Rightarrow \log_{b}(y)=x$

to get $\log_{1.05}(100)=94.387...$

How ever the answers used this rule:

$\frac {\log(100)}{\log(105)-\log(100)}\approx94.4$

which I have not encountered. Can anyone write the explicit definition of this rule for me?

$b^x=y \Rightarrow\frac {\log(y)}{\log(100b)-\log(y)}$ was my idea?

Apologies if this is a trivial question.

$$\frac {\log(100)}{\log(105)-\log(100)}=\frac {\log(100)}{\log(105/100)}=\frac {\log(100)}{\log(1.05)}=\log_{1.05}100$$
taking the logarithm of both sides we get $$t\ln(1.05)=\ln(100)$$ therefore $$t=\frac{\ln(100)}{\ln(1.05)}$$