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I was trying to solve a question on maxima-minima and I finally ended up getting this equation: $$\ln\Big(\frac{1}{x}\Big)=1$$

If I take anti-log on both sides I get $\frac{1}{x}=e$ and therefore $x=\frac{1}{e}$.

But if I expand the log as $\ln(\frac{1}{x})=\ln(1)-\ln(x)=1$, I get $x=e$.

Why are there two possible answers? Am I doing anything terribly wrong? Please correct me.

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  • $\begingroup$ ohhhhhhhh....that's the greatest careless mistake that I have done in my life.....thanks for correcting me... $\endgroup$ – Rajath Krishna R Sep 13 '13 at 13:49
  • $\begingroup$ $-\ln(x)=1$ means $\ln(x)=-1$, so the answer is still $-e$. $\endgroup$ – mau Sep 13 '13 at 13:51
  • $\begingroup$ If that is your greatest careless mistake, you have a ways to go to be competitive. $\endgroup$ – Ross Millikan Sep 13 '13 at 14:58
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Both solutions are consistent: Note, we are subtracting $\ln 1 - \ln x = 0 - \ln x = -\ln x$. So, expanding as you did in the second case, but correcting for the sign error: $$\ln\left(\frac 1x\right) = \ln (1) - \ln (x) = 1 \iff - \ln x = 1 \iff \ln x = -1 \implies x = e^{-1} = \dfrac 1e$$

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You lost a negative sign in passing from $\ln(1)-\ln(x)=1$ You should get $-\ln(x)=1$ which has solution $1/e$

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$ x = e $ actually does not satisfy the equation

$$\ln(1) - \ln(x) = 1$$

because $ \ln(1) - \ln(x) = \ln(1) - \ln(e) = 0 - 1 = -1 \neq 1$.

You are correct with the first solution $ x = \frac{1}{e} $ but it turns out to be the ''only'' solution.

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