17.) 256^ycdot 16^y-1=4^2y-22

To solve the equation \(256^{y} \cdot 16^{y-1} = 4^{2y-22}\), we start by expressing all terms as powers of 4:<br /><br />1. Express 256, 16, and 4 as powers of 4:<br /> - \(256 = 4^4\)<br /> - \(16 = 4^2\)<br /> - \(4 = 4^1\)<br /><br />2. Substitute these into the equation:<br /> \[<br /> (4^4)^y \cdot (4^2)^{y-1} = (4^1)^{2y-22}<br /> \]<br /><br />3. Simplify the exponents:<br /> \[<br /> 4^{4y} \cdot 4^{2(y-1)} = 4^{2y-22}<br /> \]<br /><br />4. Combine the exponents on the left side:<br /> \[<br /> 4^{4y + 2y - 2} = 4^{2y-22}<br /> \]<br /><br />5. Simplify the left side:<br /> \[<br /> 4^{6y - 2} = 4^{2y-22}<br /> \]<br /><br />6. Since the bases are the same, set the exponents equal:<br /> \[<br /> 6y - 2 = 2y - 22<br /> \]<br /><br />7. Solve for \(y\):<br /> \[<br /> 6y - 2y = -22 + 2<br /> \]<br /> \[<br /> 4y = -20<br /> \]<br /> \[<br /> y = -5<br /> \]<br /><br />Thus, the solution is \(y = -5\).