<s>[INST] Solve $2^{a+3}=4^{a+2}-48,\ a\in \mathbb{R}$ 
Solve $2^{a+3}=4^{a+2}-48,\ a\in \mathbb{R}$

I tried to simplify it ,
$2^{a+3}=4^{a+2}-48\\
2^{a+3}=2^{2(a+2)}-2^4\cdot 3\\
2^{2a}-2^{a-1}- 3=0\\
$
I don't know how to go from here.
This question is from chapter quadratic equations, so i think there must be hidden quadratic idea in it.
I look for a short and simple way.
I have studied maths up to $12$th grade.
 [/INST] $$2^{a+3}=4^{a+2}-48\Longleftrightarrow$$
$$48+2^{a+3}-4^{a+2}-48=0\Longleftrightarrow$$
$$-8\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$\left(2^a-2\right)=0 \vee \left(3+2^{a+1}\right)=0\Longleftrightarrow$$
$$2^a=2 \vee 3+2^{a+1}=0\Longleftrightarrow$$
$$a=\frac{\log_{10}(2)}{\log_{10}(2)} \vee 2^{a+1}=-3\Longleftrightarrow$$
$$a=1 \vee a+1=\frac{\log_{10}(-3)}{\log_{10}(2)}\Longleftrightarrow$$
$$a_1=1 \vee a_2=\frac{\log_{10}(-3)}{\log_{10}(2)}-1\Longleftrightarrow$$
$$a_1=1 \vee a_2=\frac{\ln(3)+\pi i}{\ln(2)}-1$$
And we see that $a_2$ isn't an real solution so the only real solution is $a=1$.
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