To solve the equation \( 8^{x+1} = 16^{2x-4} \), we can start by expressing both sides with base 2, since both 8 and 16 can be expressed as powers of 2:
\[ 8 = 2^3 \quad \text{and} \quad 16 = 2^4 \]
Now we rewrite the equation:
\[ (2^3)^{x+1} = (2^4)^{2x-4} \]
Using the power of a power property \((a^m)^n = a^{m \cdot n}\), we simplify both sides:
\[ 2^{3(x+1)} = 2^{4(2x-4)} \]
Now we equate the exponents since the bases are the same:
\[ 3(x+1) = 4(2x-4) \]
Now, expand both sides:
\[ 3x + 3 = 8x - 16 \]
Next, we will isolate \(x\). Start by getting all \(x\) terms on one side and the constant terms on the other side:
\[ 3 + 16 = 8x - 3x \]
This simplifies to:
\[ 19 = 5x \]
Now, solve for \(x\):
\[ x = \frac{19}{5} \]
Thus, the solution to the equation \(8^{x+1} = 16^{2x-4}\) is:
\[ \boxed{\frac{19}{5}} \]
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