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

\(4^{x}\) ↓
\((( 2^{x} )) ^2\)↓
\(y^{x} = w\)↓
\(w^{2} - w - 56 = 0\)↓
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"

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erinna erinna · Answer 2 · 2021-10-19T05:23:13-04:00

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.

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