Question

a) $log_{2}(x + 4)+log_{2}(x + 6)=3$
c) $e^{2x}-e^{x}-12 = 0$

Explanation:

Step1: Solve $\log_{2}(x + 4)+\log_{2}(x + 6)=3$

Use log - product rule $\log_aM+\log_aN=\log_a(MN)$.
$\log_{2}[(x + 4)(x + 6)]=3$
By the definition of logarithms, if $\log_{a}b=c$, then $b=a^{c}$. So $(x + 4)(x + 6)=2^{3}=8$.
Expand the left - hand side: $x^{2}+6x+4x + 24=8$, which simplifies to $x^{2}+10x+24 - 8=0$, or $x^{2}+10x + 16=0$.
Factor the quadratic equation: $(x + 2)(x+8)=0$.
Set each factor equal to zero: $x+2 = 0$ gives $x=-2$; $x + 8=0$ gives $x=-8$.
Check for domain: For $\log_{2}(x + 4)$ and $\log_{2}(x + 6)$, $x+4>0$ and $x + 6>0$, i.e., $x>-4$ and $x>-6$. So $x=-8$ is an extraneous solution. The solution of $\log_{2}(x + 4)+\log_{2}(x + 6)=3$ is $x=-2$.

Step2: Solve $e^{2x}-e^{x}-12 = 0$

Let $y = e^{x}$, then the equation becomes a quadratic equation $y^{2}-y - 12=0$.
Factor the quadratic equation: $(y - 4)(y+3)=0$.
Set each factor equal to zero: $y - 4=0$ gives $y = 4$; $y+3=0$ gives $y=-3$.
Since $y = e^{x}>0$ for all real $x$, we discard $y=-3$.
If $y = e^{x}=4$, then by the definition of the natural logarithm, $x=\ln4$.

Answer:

The solution of $\log_{2}(x + 4)+\log_{2}(x + 6)=3$ is $x=-2$; the solution of $e^{2x}-e^{x}-12 = 0$ is $x=\ln4$.