^4√4^х-3=^5√4^х+2

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:

\[ \boxed{23}. \]