To solve the equation [tex]\(16^x = 64^{x+4}\)[/tex], let's go through the steps to find the solutions.
Given equation:
[tex]\[
16^x = 64^{x+4}
\][/tex]
First, express both sides of the equation with a common base. Notice that [tex]\(16\)[/tex] and [tex]\(64\)[/tex] can both be written as powers of [tex]\(2\)[/tex]:
[tex]\[
16 = 2^4
\][/tex]
[tex]\[
64 = 2^6
\][/tex]
Therefore, rewrite the equation as:
[tex]\[
(2^4)^x = (2^6)^{x+4}
\][/tex]
This simplifies to:
[tex]\[
2^{4x} = 2^{6(x+4)}
\][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[
4x = 6(x + 4)
\][/tex]
Expand and simplify the equation:
[tex]\[
4x = 6x + 24
\][/tex]
Subtract [tex]\(6x\)[/tex] from both sides:
[tex]\[
4x - 6x = 24
\][/tex]
[tex]\[
-2x = 24
\][/tex]
Divide both sides by [tex]\(-2\)[/tex]:
[tex]\[
x = -12
\][/tex]
Thus, [tex]\(x = -12\)[/tex] is indeed one of the solutions.
The second solution seems to be more complex and contains a logarithmic term. Therefore, there is another solution:
[tex]\[
x = \frac{-\log(4096) + i\pi}{\log(2)}
\][/tex]
However, only the real solution from our equation is [tex]\(x = -12\)[/tex].
Therefore, the only real value of [tex]\(x\)[/tex] that is a solution to the equation [tex]\(16^x = 64^{x+4}\)[/tex] is:
[tex]\[
x = -12
\][/tex]
