state the values of x for which the following identity is true

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

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- Apr 4th 2010, 11:30 PMugieLogarithm 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 PMICanFly
Hint: Log(xy)=log x + log y and If we have log a, a must be >0

- Apr 4th 2010, 11:38 PMundefined
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 PMStroodle
$\displaystyle log(x+3)+log(x+4)=log\left [(x+3)(x+4)\right ]=log(x^2+7x+12)$

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