# Thread: help with log problem pls?

1. ## help with log problem pls?

Hi all i need urgent help on a log question thats on our review, i think i might have th e right steps down butnot sure

heres what i have tofar

log x +log(x+15)=2

log x[x(x+15)]=2

x(x+15)=2^10

then x^2+15x=1024

Then x^2+15x-1024=0

to solve and get 2 answers of x

thing is Im not able t go any furhter cant get any answers of 1024 that could equal 15, and same for quadratic forumla

pls help as i need to despr. now this

thanks

2. Originally Posted by billythekid

to solve and get 2 answers of x
Put those answers into the original equation and see if they solve it. If they both do then both of them are the answers. If one does and one does not then the one that does is the solution. Otherwise it has no solutions.

3. Originally Posted by billythekid
Hi all i need urgent help on a log question thats on our review, i think i might have th e right steps down butnot sure

heres what i have tofar

log x +log(x+15)=2

log x[x(x+15)]=2

x(x+15)=2^10

then x^2+15x=1024

Then x^2+15x-1024=0

to solve and get 2 answers of x

thing is Im not able t go any furhter cant get any answers of 1024 that could equal 15, and same for quadratic forumla
Do you know I went through a full derivation of this and had concluded there were no solutions? Then I finally saw what went wrong:
$log_{10}(x) + log_{10}(x + 15) = 2$

$log_{10}(x(x + 15)) = 2$

$x^2 + 15x = 10^2$ <---

$x^2 + 15x - 100 = 0$

$(x - 5)(x + 20) = 0$

Thus x = 5 or x = -20. But x can't be negative else $log_{10}(x)$ won't exist.

Thus x = 5.

-Dan