# Solve $3^{1-x}=2$

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?

• "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 0:49
• @AdamRubinson Thanks - I'll do that, appreciate the useful advice on how to reach the solution – user133935 Nov 29 '20 at 0:51

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 $$\log_{b}x = a = \frac{\ln x}{\ln b}.$$

• Fantastic, thank you - I will look this up – user133935 Nov 29 '20 at 0:48

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

• Fantastic, thank you - I will look this up – user133935 Nov 29 '20 at 0:48
• @user133935: This answer is wrong for $e=1$. – user21820 Nov 29 '20 at 9:58
• The base of natural logarithms $e\ne1$ – J. W. Tanner Nov 29 '20 at 13:13

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 $$log_3\frac{3}{2}$$.