Exponential equation question (can't solve)

Question

I came upon this question on a website:

Find all the real solutions to $4^x-2^x=56$.

I've tried to factor the expression: $2^x(2^x-1)=56$, but I don't know how to proceed. How can I solve this?

Answer 1

Hint: You can write this equation as $$(2^x)^2-(2^x)-56 = 0$$ $$(2^x-8)(2^x + 7)=0$$ Can you proceed from here?

Bonus: Find the full set of complex solutions as well!

Answer 2

$4^{x}=2^{2x}=(2^{x})^{2}$. If you put $y=2^{x}$ the equation becomes $y^{2}-y=56$ Can you continue?

Answer 3

$2^{2x}-2^x=56\Rightarrow 2^x(2^x-1)=2^3.(2^3-1)$

It is clearly visible here that $2^x=2^3\Rightarrow x=3$ which can also be found by solving the quadratic equation as given in the answer by Kavi Rama Murthy.

Answer 4

We have $$4^x-2^x+0.25=56.25$$ or $$(2^x-0.5)^2=7.5^2.$$ Can you end it now?

I got $x=3.$

Answer 5

I've got it now. Here is my answer.

First, we make the bases of the powers of $x$ equal. We notice that $4=2^2$, so: $$4^x-2^x=(2^2)^x-2^x=2^{2x}-2^x=56.$$ Now that we have done this, we can factor our expression. $$2^x(2^x-1)=56.$$

We can gave $2^x=m$ to get a quadratic expression: $$m(m-1)=56.$$ Now, we expand and move all terms to the left-hand side. $$m^2-m-56=0.$$ Factoring, we get: $$(m-8)(m+7)=0$$$$\therefore m=8, m=-7.$$

Since we have used the substitution $2^x=m$, and we are working out $x$, then we have the equations: $$2^x=8, 2^x=-7.$$ We know that $8=2^3$, so one value of $x$ is $3$. Now, we can look at $2^x=-7$ But no matter what real value of $x$ we use, we cannot get a negative output! So -7 cannot be a solution.

ANSWER: The solution to the equation $4^x - 2^x = 56$ is $x=3$.