Question

Solution of $1+3^{\frac{x}{2}}=2^x$ is

• A. $x=1$
• B. $x=2$
• C. $x=0$
• D. None of these

Answer 1

The correct option is B

$x=2$

Explanation for the correct answer:

Simplifying the given expression as

$$\begin{gathered} \Rightarrow 1+3^{\frac{x}{2}}=2^x \\ \Rightarrow {\left(1\right)}^x+{\left(3\right)}^{\frac{x}{2}}=2^x\left[\because 1^x=1\right] \\ \Rightarrow {\left(1\right)}^x+{\left(3\right)}^{\frac{1}{2}.x}=2^x \\ \Rightarrow {\left(1\right)}^x+{\left(\sqrt{3}\right)}^x=2^x\mathbf{}\left[\because {\left(3\right)}^{\frac{1}{2}}=\sqrt{3}\right] \end{gathered}$$

Dividing by $2^x$ on both sides, we get

$$\begin{gathered} {\left(\frac{1}{2}\right)}^x+{\left(\frac{\sqrt{3}}{2}\right)}^x=1 \\ \Rightarrow {\left(\cos \left(\frac{\pi }{3}\right)\right)}^x+{\left(\sin \left(\frac{\pi }{3}\right)\right)}^x=1\mathbf{}\left[\because \cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\text{and}\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}\right] \end{gathered}$$

But we know that ${\cos }^{2}x+{\sin }^{2}x=1$, $\forall x\in \mathrm{ℝ}$.

Thus on comparing, we get $x=2$.

Therefore, the correct answer is option (B).