# Logarithms melting my brain

So I've got an inequality: $\ln(2x-5) > \ln(7-2x)$ and I attempt to solve by doing the following:

$$\frac{\ln(2x)}{\ln(5)} > \frac{\ln(7)}{\ln(2x)}$$ $$\Rightarrow \ln(2x) \cdot \ln(2x) > \ln(7) \cdot \ln(5)$$ $$\Rightarrow \ln^2(2x) < \ln(7) \cdot \ln(5)$$ $$\Rightarrow 2x < \ln(7) \cdot \ln(5) \cdot e^2$$ I thought I could multiply by $e^2$ to get rid of $\ln^2$ but I guess not...I thought they were inverses??? So anyways, my solution seems WAY off as the solution to the problem is: $3 < x < \frac{7}{2}$.

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$\ln(a-b)\ne \ln a/\ln b$. Exponentiate both sides first. –  David Mitra Feb 22 '14 at 11:48
$$\frac ab>\frac cd\implies k\cdot\frac ab>k\cdot\frac cd$$ if $k>0$ –  lab bhattacharjee Feb 22 '14 at 11:49
Oh! Hold on, let me see if I can work this.. –  user3200098 Feb 22 '14 at 11:50
Yes, exponentiation is the inverse of taking logarithms but exponentiation doesn't mean multiplying by $e$. –  David Richerby Feb 22 '14 at 13:16
Quite a nice title! :) –  Sawarnik Feb 22 '14 at 15:14

You applied an identity for logarithms in the wrong direction. What we have is $$\ln(a/b) = \ln(a)-\ln(b)$$ and not the other way around: $\ln(a-b)\ne \ln(a)/\ln(b)$.

In fact, we don't have formula for $\ln(a-b)$.

But here what we only need to say is that the logarithm $\ln$ is a strictly increasing function throughout its domain (!!!), so for any numbers $A,B$ in the domain of $\ln$, we have $\ln(A)>\ln(B)$ iff $A>B\$ (and this is simply because $X>Y\ \iff\ e^X>e^Y$).

So, now we then only need to solve $2x-5\ >\ 7-2x$, and care about only those values $x$ for which both $\ln(2x-5)$ and $\ln(7-2x)$ are defined.

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That's a clever way of going about it. But I thought $e^x$ and the natural log were inverses so that when I multiply a natural log of $n$ by $e^n$ I'd get $n$ but that doesn't seem to work. How are they inverses and how could I use that to my advantage when I have $ln(x) = 5$ for example. –  user3200098 Feb 22 '14 at 11:56
Inverses: $e^{\ln(x)}=x$ and $\ln(e^y)=y$. –  Berci Feb 22 '14 at 11:57
Ah I see. Thank you. –  user3200098 Feb 22 '14 at 11:57
Where does the $x<\frac{7}{2}$ come from in the answer in the book? –  user3200098 Feb 22 '14 at 12:02
Of course, they are not defined below 0! Thank you again! Best answer chosen. –  user3200098 Feb 22 '14 at 12:04