# Logarithm problem

• Apr 4th 2010, 11:30 PM
ugie
Logarithm problem
state the values of x for which the following identity is true
log(x+3) + log(x+4) = log(x^2 + 7x + 12)
• Apr 4th 2010, 11:38 PM
ICanFly
Hint: Log(xy)=log x + log y and If we have log a, a must be >0
• Apr 4th 2010, 11:38 PM
undefined
Quote:

Originally Posted by ugie
state the values of x for which the following identity is true
log(x+3) + log(x+4) = log(x^2 + 7x + 12)

This is how I would proceed:

We rely on the principle that log(xy) = log(x) + log(y).

LHS means left-hand-side, RHS right-hand-side.

LHS: log(x+3) + log(x+4) = log((x+3)(x+4)) = log(x^2 + 7x + 12)

Since this is the same as the RHS, all we need to do is ensure that the quantities (x+3), (x+4) and (x^2 + 7x + 12) are positive -- otherwise, taking their log is undefined.

So we are solving the simultaneous inequalities

x + 3 > 0
x + 4 > 0
(x+3)(x+4) > 0
• Apr 4th 2010, 11:39 PM
Stroodle
$\displaystyle log(x+3)+log(x+4)=log\left [(x+3)(x+4)\right ]=log(x^2+7x+12)$

So $\displaystyle x\in(-3,\infty)$, as you can't take the log of a negative value.