Math Help - Two Logarithmic Function Problems

1. Two Logarithmic Function Problems

I'm having trouble solving these two problems:

1.) $log_2(x+7) + log_2(x) = 3$

2.) $e^x + 2 = 8e^{-x}$

2. Ignore the solution to the first question, there's a mistake in it.
log_c a + log_c b = log_c a.b
So,
log_2(x+7) + log_2(x) = 3
log_2{x^2 + 7} = 3
x^2 + 7 = 2^3 = 8
x^2 = 1
x={-1,1}
Don't forget to check the results because x+7 > 0 and x>0.
Ah, x = -1 doesn't satisfy x>0. So we remove it from the solution set.
x=1

----------------------------------------------

$e^x + 2 = 8e^{-x}$

$e^x + 2 = \frac{8}{e^x}$

Let $a=e^x$

$a + 2 = \frac{8}{a}$

Multiply both sides by a,

$a^2 + 2a = 8$

$a^2 + 2a - 8 = 0$

$(a+4)(a-2)=0$

$a=\{ -4 , 2 \}$

Put these a values back in $a=e^x$,

$-4 = e^x$

$x = \ln -4$ (Not real!)
(We exclude this because it's impossible to have such real x)

$2 = e^x$

$x = \ln 2$

3. The rule should be: $\log_{c}a + \log_{c}b = log_{c}(ab)$

$\log_{2}(x + 7) + \log_{2}(x) = 3$
$\log_{2}\left[(x + 7)(x)\right] = 3$
$\log_{2}\left(x^{2} + 7x\right) = 3$
$x^{2} + 7x = 2^{3}$
$x^{2} + 7x = 8$

4. Sorry, o_O is right. Ignore my solution to the first problem, I didn't multiply 7 by x. But the rest is similar, find the roots, check them for $x+7>0$ and $x>0$.