How to solve the equation $\log_4 (2^x + 48) = x - 1$

Question

Please assist in solving for $x$ given the following equation:

$$\log_4(2^x + 48) = x-1$$

I understand that we can we write this as

$$2^x + 48 = 4^{x-1}$$

which can be written as

$$2^x + 48 = 2^{2x-2}$$

I however get stuck when here.

Answer 1

Your last equation can be written as $$2^x + 48 = \frac{1}{4}(2^x)^2$$ and this is an easy equation in $y = 2^x$.

EDIT : the discriminant of the equation $y^2 - 4y - 4*48 = 0$ is $28^2$, so the solutions are $-12$ and $16$. Hence, the only solution of your initial equation is $x = 4$.

Answer 2

Raise both sides to the power of $4$ to get:

$$2^x + 48 = 4^{x-1} = 4^x 4^{-1} = \frac{1}{4} \cdot 4^x$$

Now $4^x = 2^x \cdot 2^x = (2^x)^2$, and set $u = 2^x$ to form a quadratic.