# How do you solve ln(x-3)+ln(x+4)=3ln2?

Aug 20, 2015

I found $x = 4$

#### Explanation:

You can first use the property of logs:
$\ln x + \ln y = \ln \left(x \cdot y\right)$
so:
$\ln \left(x - 3\right) + \ln \left(x + 4\right) = \ln \left(\left(x - 3\right) \left(x + 4\right)\right)$
then the other property:
$a \ln x = \ln {x}^{a}$
so you get:
$\ln \left(\left(x - 3\right) \left(x + 4\right)\right) = \ln {2}^{3}$
Now take the exponential of both sides:
${e}^{\ln \left(\left(x - 3\right) \left(x + 4\right)\right)} = {e}^{\ln {2}^{3}}$
cancelling $e$ and $\ln$ you get:
$\left(x - 3\right) \left(x + 4\right) = {2}^{3}$
${x}^{2} + 4 x - 3 x - 12 = 8$
${x}^{2} + x - 20 = 0$
${x}_{1 , 2} = \frac{- 1 \pm \sqrt{1 + 80}}{2}$
${x}_{1 , 2} = \frac{- 1 \pm 9}{2}$
Two solutions:
${x}_{1} = - 5$
${x}_{2} = 4$
Try the two into the original equation and you'll find that the negative does not work.

Jan 12, 2016

Minutely different way of talking this problem

#### Explanation:

Addition of logs is the result of multiplication of the source numbers

So $\textcolor{b r o w n}{\ln \left(x - 3\right) + \ln \left(x + 4\right)} \textcolor{b l u e}{\to \ln \left(\textcolor{w h i t e}{\ldots} \left(x + 3\right) \left(x + 4\right) \textcolor{w h i t e}{\ldots}\right)}$

Multiplication of a log is the result of the source number being raised to a power

So $\textcolor{b r o w n}{3 \ln \left(2\right)} \textcolor{b l u e}{\to \ln \left({2}^{3}\right)}$

Putting it all together we have

color(brown)(ln(color(blue)((x+3)(x+4)))=ln(color(blue)(2^3))

$\textcolor{g r e e n}{\text{As both side are direct logs of each other we can simply forget}}$$\textcolor{g r e e n}{\text{about the logs and write:}}$

$\left(x + 3\right) \left(x + 4\right) = {2}^{3}$

Now it can be solved as in Gio's solution