Find all real solutions x to

(4^x)-(2^x)=56. Please explain in steps! Thank you.

(4^x)-(2^x)=56

(2^2)^x - 2^x - 56 = 0
(2^x)^2 - 2^x - 56 = 0
let 2^x = t , we get
t^2 - t - 56 = 0
(t-8)(t+7) = 0
t = 8 or t = -7

if t = -7
2^x = -7 , no such x exists
or
if t = 8
2^x = 8 -----> x = 3

To find all real solutions to the equation (4^x) - (2^x) = 56, we can start by substituting variables to simplify the equation. Let's say y = (2^x).

This substitution allows us to rewrite the equation as follows:

(4^x) - y = 56

Next, we can substitute another variable for (4^x). Let's say z = (4^x).

The equation can now be written as:

z - y = 56

From the given equation, we can see that (4^x) can be expressed as y + 56.

Substituting z with y + 56, we have:

y + 56 - y = 56

The 'y' variables cancel each other out, leaving us with the equation:

56 = 56

Since this equation is always true for any value of 'x', it means that there are infinite real solutions for the equation (4^x) - (2^x) = 56.

Therefore, any real value of 'x' will satisfy the equation.