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I am working on an excel project and had a few questions. I need to solve $\log(x)/\log(2)=-0.145509439$. I already know the answer, which is $0.90406$, but I do not know how to solve the equation.

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  • $\begingroup$ Just cross multiply ;) $\endgroup$ – BigM Feb 20 '15 at 1:04
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You need to use the change of base formula, which gives you $\log_2(x) = -0.145509439$ and then you know that $2^{-0.145509439} = x$ and arithmetic gives you $x=.90406$

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Multiply both sides by $\log(2)$, and then exponentiate to turn the $\log(x)$ into just $x$.

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  • $\begingroup$ So basically I get down to log(x)=-0.0438027058. What do I exponentiate? Sorry haven't done this stuff in years. $\endgroup$ – Brennan Feb 20 '15 at 1:09
  • $\begingroup$ @Brennan $\log(x)$ is the inverse function of $e^x$. In other words, if $\log(x)=y$, then $x=e^y$. (Also: $\ln(e^x)=e^{\ln x}=x$.) $\endgroup$ – Akiva Weinberger Feb 20 '15 at 1:16
  • $\begingroup$ OK so I did that and got the wrong answer for some reason. I multiplied by log(2) on each side and got -0.0438027058. Then basically I have log(x)=-0.0438027058. Then I do e^-0.0438027058. Which gives me 95.71%, the correct answer is 90.406%. $\endgroup$ – Brennan Feb 20 '15 at 1:26
  • $\begingroup$ In this case I think the log is in base 10, not base e. $\endgroup$ – Esteemator Feb 20 '15 at 9:37
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So we have $\frac{logx}{log2}=-0.145509439$ multiply both sides by $log2$ to have $logx=-0.100859...$ and then apply exponential function to both sides. $e^{logx}=e^{-0.100859...}$ and you should know that $e^{logx}=x$ hence $x=e^{-0.100859...}=0.9046...$

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