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Solve

$$\ln (2 x-5)>\ln (7-2 x)$$

The answer is given as $$3<x<7/2$$

This is what I have done $$\ln (2 x-5)-\ln (7-2 x)>0$$

$$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$

However I am not able to understand how to get to the answer provided.

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    $\begingroup$ If $\ln(y)>0$ then $y>1$ $\endgroup$
    – rlartiga
    Aug 25, 2014 at 15:37

3 Answers 3

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Good job, you're almost there:

$$\ln \left(\frac{2 x-5}{7-2 x}\right)>0 \\ \frac{2 x-5}{7-2 x}>\underbrace{e^0}_{1} \\ 2 x-5>7-2x \\ 4x>12 \\ x>3$$

That takes care of the first part, but also recall that the parameter of $\ln$ has to be greater than $0$ ($e$ to any power will not give you a number less than or equal to $0$). In other words, you also have to say this:

$$\begin{cases}2 x-5>0\\7-2 x>0\end{cases}$$

Solved: $$\begin{cases}x>\frac52\\x<\frac72\end{cases}$$

$x$ is indeed greater than $\frac{5}{2}$, as we found out, but it must also be geater than $3$. Hence we can ignore that because we have $x>3$. Now we have an upper bound, of $x<\frac72$, because after that point the answer won't be defined. So the correct answer is $$\therefore 3<x<\frac72$$

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Hint

If $$\ln \left(\frac{2 x-5}{7-2 x}\right)>0$$ it implies that $$\frac{2 x-5}{7-2 x}>1$$ But take care : the logarithm is such that its argument must be positive.

I am sure that you can take from here.

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Exponentiation is a one to one, increasing function so:

$\ln (2x-5) > \ln (7-2x)\\ e^{\ln (2x-5)} > e^{\ln (7-2x)}\\ 2x-5>7x-2\\ ...\\ x>3$

However there are the domains to consider as well:

$2x-5>0$ gives $x>\frac{5}{2}$ and $7-2x>0$ gives $x<7/2$. We have to put all of the inequalities $x>3$ and $x>\frac{5}{2}$ and $x<\frac{7}{2}$ Together. This gives the answer $3<x<\frac{7}{2}$

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  • $\begingroup$ One-to-one is implied by increasing. $\endgroup$ Aug 25, 2014 at 15:47
  • $\begingroup$ @GammaFunction for continuous functions. $\endgroup$
    – Alice Ryhl
    Aug 25, 2014 at 15:49
  • $\begingroup$ One-to-one is implied by strictly increasing. See en.wikipedia.org/wiki/Monotonic_function $\endgroup$ Aug 25, 2014 at 15:50
  • $\begingroup$ @Darksonn If $D \subseteq \mathbb{R}$ and $f:D \to \mathbb{R}$ is increasing, then $x,y \in D$ and $x > y$ implies $f(x)>f(y)$. Thus if $x \neq y$, we have $f(x) \neq f(y)$. $\endgroup$ Aug 25, 2014 at 15:57

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