# Math Help - Log Equation

1. ## Log Equation

$
\ln x + \ln (x+2) = 4
$

The answer is:
$
x = -1 + \sqrt{1 + e^4}
$

so far I've done:

$
log_e x + log_e (x+2) = e^4
$

$
log_e x^2+2x = e^4
$

Stuck.

2. Originally Posted by desiderius1
$
\ln x + \ln (x+2) = 4
$

The answer is:
$
x = -1 + \sqrt{1 + e^4}
$

so far I've done:

$
log_e x + log_e (x+2) = e^4
$
No it's not

$
log_e x^2+2x = e^4
$
In the future please put brackets around x^2 + 2x

Stuck.

$\ln x + \ln (x+2) = 4$

$\ln | x(x+2)| = 4$

$x^2 + 2x = e^4$

$(x+1)^2 = e^4 + 1$

$x+1 = \sqrt {e^4 + 1}$

$x = -1 + \sqrt{1+e^4}$

3. Originally Posted by desiderius1
$
\ln x + \ln (x+2) = 4
$

The answer is:
$
x = -1 + \sqrt{1 + e^4}
$

so far I've done:

$
log_e x + log_e (x+2) = e^4
$
That's NOT your orignal equation!

$
log_e x^2+2x = e^4
$

Stuck.