I'm having trouble solving these two problems:
1.) $\displaystyle log_2(x+7) + log_2(x) = 3$
2.) $\displaystyle e^x + 2 = 8e^{-x}$
Your assistance is much appreciated.
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
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$\displaystyle e^x + 2 = 8e^{-x}$
$\displaystyle e^x + 2 = \frac{8}{e^x}$
Let $\displaystyle a=e^x$
$\displaystyle a + 2 = \frac{8}{a}$
Multiply both sides by a,
$\displaystyle a^2 + 2a = 8$
$\displaystyle a^2 + 2a - 8 = 0$
$\displaystyle (a+4)(a-2)=0$
$\displaystyle a=\{ -4 , 2 \}$
Put these a values back in $\displaystyle a=e^x$,
$\displaystyle -4 = e^x$
$\displaystyle x = \ln -4$ (Not real!)
(We exclude this because it's impossible to have such real x)
$\displaystyle 2 = e^x$
$\displaystyle x = \ln 2$
The rule should be: $\displaystyle \log_{c}a + \log_{c}b = log_{c}(ab)$
$\displaystyle \log_{2}(x + 7) + \log_{2}(x) = 3$
$\displaystyle \log_{2}\left[(x + 7)(x)\right] = 3$
$\displaystyle \log_{2}\left(x^{2} + 7x\right) = 3$
$\displaystyle x^{2} + 7x = 2^{3}$
$\displaystyle x^{2} + 7x = 8$
Thus, a simple quadratic.