Solve $2^{a+3}=4^{a+2}-48,\ a\in \mathbb{R}$

Question

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.

Answer 1

HINT :

We have $$(2^a)^2-\frac 12\cdot 2^a-3=0.$$ Then, set $t=2^a$.

Answer 2

Hint: Let $x=2^a$, and see what you obtain in terms of $x$.

Answer 3

You're almost there! Now introduce $a = \log_2(b)$ to convert your problem into a quadratic of $b$. You can use the quadratic formula to solve the equation and then cast your answer to one in terms of $a$.

Answer 4

Let $t=2^a$ to have $$t^2 -\frac{t}{2} -3=0$$ and use the quadratic formula.

Answer 5

$$2^{a+3}=4^{a+2}-48\Longleftrightarrow$$ $$48+2^{a+3}-4^{a+2}-48=0\Longleftrightarrow$$ $$-8\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$ $$\left(2^a-2\right)\left(3+2^{a+1}\right)=0\Longleftrightarrow$$ $$\left(2^a-2\right)=0 \vee \left(3+2^{a+1}\right)=0\Longleftrightarrow$$

$$2^a=2 \vee 3+2^{a+1}=0\Longleftrightarrow$$ $$a=\frac{\log_{10}(2)}{\log_{10}(2)} \vee 2^{a+1}=-3\Longleftrightarrow$$ $$a=1 \vee a+1=\frac{\log_{10}(-3)}{\log_{10}(2)}\Longleftrightarrow$$ $$a_1=1 \vee a_2=\frac{\log_{10}(-3)}{\log_{10}(2)}-1\Longleftrightarrow$$ $$a_1=1 \vee a_2=\frac{\ln(3)+\pi i}{\ln(2)}-1$$

And we see that $a_2$ isn't an real solution so the only real solution is $a=1$.