To solve the equation ( \sqrt[4]{4^{x-3}} = \sqrt[5]{4^{x+2}} ), we can first rewrite the roots in terms of exponents.
The fourth root can be expressed as:
[
\sqrt[4]{4^{x-3}} = (4^{x-3})^{1/4} = 4^{\frac{x-3}{4}}.
]
The fifth root can be expressed as:
[
\sqrt[5]{4^{x+2}} = (4^{x+2})^{1/5} = 4^{\frac{x+2}{5}}.
]
Now, substituting these into the original equation, we get:
[
4^{\frac{x-3}{4}} = 4^{\frac{x+2}{5}}.
]
Since the bases on both sides are the same (both are 4), we can set the exponents equal to each other:
[
\frac{x-3}{4} = \frac{x+2}{5}.
]
To eliminate the fractions, we can cross-multiply:
[
5(x - 3) = 4(x + 2).
]
Expanding both sides:
[
5x - 15 = 4x + 8.
]
Now, isolate ( x ) by subtracting ( 4x ) from both sides:
[
5x - 4x - 15 = 8,
]
which simplifies to:
[
x - 15 = 8.
]
Add 15 to both sides:
[
x = 23.
]
Thus, the solution to the equation is:
[
\boxed{23}.
]
