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I've been trying to solve a problem:

$3^{1-x}=2$

I converted this to a log as: $\log_{3}{2} = (1-x)$

But I couldn't see how to progress from there. Having put it into a solving team, it suggests that it can be translated to: $\left(1-x\right)\ln \left(3\right)=\ln \left(2\right)$

...but I can't see why this makes sense. Could anyone point me in the right direction?

user21820
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user133935
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    "But I couldn't see how to progress from there". You have an equation with one unknown, namely $x$. Re-arrange to make $x$ the subject of the equation [ Hint: the brackets don't serve a purpose in the equation:$log_{3}{2} = (1-x)\ $] – Adam Rubinson Nov 29 '20 at 00:49
  • @AdamRubinson Thanks - I'll do that, appreciate the useful advice on how to reach the solution – user133935 Nov 29 '20 at 00:51

3 Answers3

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This is due to the change of base formula, $$\log_{b}x = \frac{\ln x}{\ln b}.$$

This is because if you have $\log_{b}x = a$, then you can rewrite it as $x = b^{a}$. Taking the natural logarithm of each side gives you $\ln x = \ln (b^a)$, which can be rewritten as $\ln x = a\ln b$, which upon dividing both sides by $\ln b$ gives you $a = \frac{\ln x}{\ln b}$. Hence, you have $$\log_{b}x = a = \frac{\ln x}{\ln b}.$$

DMcMor
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Use the property of logarithms that $\log_b x=\dfrac {\log_e x}{\log_e b}$.

J. W. Tanner
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If you mean to find the value of $x$ you almos have it. Another option is to write $3^{(1-x)}=2$ as $\frac{3}{3^x}=2$ which is equivalent to $\frac{3}{2}=3^x$ and using $log_3$ the value of $x$ is $log_3\frac{3}{2}$.

Navareño
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