Question

find the value of x (square root of 32) ÷ 2^x+1 = 1

Answer 1

$$\begin{gathered} \frac{\sqrt{32}}{2^{\mathrm{x}+1}}=1 \\ \Rightarrow \frac{\sqrt{2^5}}{2^{\mathrm{x}+1}}=2^0 \\ \Rightarrow \frac{2^{5/2}}{2^{\mathrm{x}+1}}=2^0 \\ \Rightarrow 2^{\frac{5}{2}-\mathrm{x}-1}=2^0 \\ \Rightarrow \frac{5}{2}-\mathrm{x}-1=0 \\ \Rightarrow \frac{3}{2}-\mathrm{x}=0 \\ \Rightarrow \boxed{\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{3}}{\mathbf{2}}} \end{gathered}$$