Question

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

Answer 1

$4^{x}$ ↓
 $(( 2^{x} )) ^2$↓
$y^{x} = w$↓
$w^{2} - w - 56 = 0$↓
w - 8  &  w + 7↓
w(1) = 8 → w(1) = $2^{x1} = 8$ x(1) = 3↓
w(2) = -7 → w(2) = $2^{x2} = -7$
So, therefore the value of "x" would have to be "3" 

Answer 2

The value of $x$ is 3.

Given:

The given equation is:

$4^x-2^x=56$

To find:

The value of $x$.

Explanation:

The given equation can be rewritten as:

$(2^2)^x-2^x-56=0$

$(2^x)^2-2^x-56=0$

Substitute $2^x=t$ in the above equation.

$t^2-t-56=0$

Split the middle term.

$t^2-8t+7t-56=0$

$t(t-8)+7(t-8)=0$

$(t-8)(t+7)=0$

$t=8,-7$

Now,

$2^x=8$

$2^x=2^3$

$x=3$

And,

$2^x=-7$

This statement is not true for any real number $x$.

Therefore, the only real value of $x$ is 3.

Learn more:

https://brainly.com/question/19956884