# Two possible answers for x

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.

• ohhhhhhhh....that's the greatest careless mistake that I have done in my life.....thanks for correcting me... – Rajath Krishna R Sep 13 '13 at 13:49
• $-\ln(x)=1$ means $\ln(x)=-1$, so the answer is still $-e$. – mau Sep 13 '13 at 13:51
• If that is your greatest careless mistake, you have a ways to go to be competitive. – Ross Millikan Sep 13 '13 at 14:58

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: $$\ln\left(\frac 1x\right) = \ln (1) - \ln (x) = 1 \iff - \ln x = 1 \iff \ln x = -1 \implies x = e^{-1} = \dfrac 1e$$
You lost a negative sign in passing from $\ln(1)-\ln(x)=1$ You should get $-\ln(x)=1$ which has solution $1/e$
$$x = e$$ 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.