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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.

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Your idea and result are correct.

What the answer uses is something equivalent

$$\frac {\log(100)}{\log(105)-\log(100)}=\frac {\log(100)}{\log(105/100)}=\frac {\log(100)}{\log(1.05)}=\log_{1.05}100$$

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

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  • $\begingroup$ yes thank you it is to late for me $\endgroup$ – Dr. Sonnhard Graubner Jul 11 '17 at 20:18
  • $\begingroup$ thank you randomgirl (nice name) $\endgroup$ – Dr. Sonnhard Graubner Jul 11 '17 at 20:19

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