Solve for \( \mathrm{x}: 4^{x+1}=45 \) \( \begin{array}{lll}\text { a. } 1.746 & \text { b. } & 3.746 \\ \text { c. }-0.636 & \text { d. } & 1.364\end{array} \)

Solve the equation \( 4^{x+1}=45 \). Solve the equation by following steps: - step0: Solve for \(x\): \(4^{x+1}=45\) - step1: Take the logarithm of both sides: \(\log_{4}{\left(4^{x+1}\right)}=\log_{4}{\left(45\right)}\) - step2: Evaluate the logarithm: \(x+1=\log_{4}{\left(45\right)}\) - step3: Move the constant to the right side: \(x=\log_{4}{\left(45\right)}-1\) - step4: Simplify: \(x=\frac{1}{2}\log_{2}{\left(45\right)}-1\) - step5: Simplify: \(x=\frac{\log_{2}{\left(45\right)}-2}{2}\) The solution to the equation \(4^{x+1}=45\) is \(x=\frac{\log_{2}(45)-2}{2}\). Now, we need to calculate the value of \(x\) using the given solution. Let's calculate the value of \(x\): \[ x = \frac{\log_{2}(45)-2}{2} \] Using a calculator, we find that: \[ x \approx 1.746 \] Therefore, the correct answer is: \[ \text{a. } 1.746 \]