Question

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

Answer 1

I found `x=4`

Explanation:

You can first use the property of logs:
`lnx+lny=ln(x*y)`
so:
`ln(x-3)+ln(x+4)=ln((x-3)(x+4))`
then the other property:
`alnx=lnx^a`
so you get:
`ln((x-3)(x+4))=ln2^3`
Now take the exponential of both sides:
`e^(ln((x-3)(x+4)))=e^(ln2^3)`
cancelling `e` and `ln` you get:
`(x-3)(x+4)=2^3`
`x^2+4x-3x-12=8`
`x^2+x-20=0`
Using the Quadratic Formula:
`x_(1,2)=(-1+-sqrt(1+80))/2`
`x_(1,2)=(-1+-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.

Answer 2

Minutely different way of talking this problem

Explanation:

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

So `color(brown)(ln(x-3)+ln(x+4))color(blue)( -> ln(color(white)(...)(x+3)(x+4)color(white)(...)))`

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

So `color(brown)(3ln(2))color(blue)(-> ln(2^3))`

Putting it all together we have

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

`color(green)("As both side are direct logs of each other we can simply forget")` `color(green)("about the logs and write:")`

`(x+3)(x+4)=2^3`

Now it can be solved as in Gio's solution