I wanted to change the base of this logarithm $\log_432$ from 4 to 2 so it could be easier to solve. Using some properties I get 2.5 as a result. However I was told that if I use $\frac{\ln 32}{\ln 4}$ I get the same result. Can someone explain me how this ln works?
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1It’s the change of base formula it is: $\log_b{x} = \frac{\log_a{x}}{\log_a{b}}$ – Cotton Headed Ninnymuggins Nov 03 '20 at 21:16
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In general$$y:=\log_ax\implies b^{y\log_b a}=a^y=x\implies y\log_b a=\log_bx\implies y=\tfrac{\log_bx}{\log_ba},$$whether you use $b=2$ or $b=e$.
J.G.
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I can never answer fast enough to make it worth answering anything lol. The 0.001% of the questions I know the answer to... I’m too slow. – Cotton Headed Ninnymuggins Nov 03 '20 at 21:21
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$\log_432$ is the number $x$ such that $4^x=32$.
Taking the natural logarithm at both sides, we get
$$\ln(4^x)=x·\ln(4)=\ln(32)$$
Finally, $$x=\frac{\ln(32)}{\ln(4)}$$
Jaume Oliver Lafont
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