Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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: $3 < x < \frac{7}{2}$.

share|improve this question
3  
$\ln(a-b)\ne \ln a/\ln b$. Exponentiate both sides first. –  David Mitra Feb 22 at 11:48
    
$$\frac ab>\frac cd\implies k\cdot\frac ab>k\cdot\frac cd$$ if $k>0$ –  lab bhattacharjee Feb 22 at 11:49
    
Oh! Hold on, let me see if I can work this.. –  user3200098 Feb 22 at 11:50
    
Yes, exponentiation is the inverse of taking logarithms but exponentiation doesn't mean multiplying by $e$. –  David Richerby Feb 22 at 13:16
    
Quite a nice title! :) –  Sawarnik Feb 22 at 15:14

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
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 at 11:56
    
Inverses: $e^{\ln(x)}=x$ and $\ln(e^y)=y$. –  Berci Feb 22 at 11:57
    
Ah I see. Thank you. –  user3200098 Feb 22 at 11:57
    
Where does the $x<\frac{7}{2}$ come from in the answer in the book? –  user3200098 Feb 22 at 12:02
    
Of course, they are not defined below 0! Thank you again! Best answer chosen. –  user3200098 Feb 22 at 12:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.