Solve $1 + 3^{x/2} = 2^x$ 
Solve $1 + 3^{x/2} = 2^x$.

I tried to replace $3^{x/2}$ with $2^{\log_2(3) * x/2}$
Also I tried to use derivatives, saying that $x=2$ is a root, and then comparing the derivatives of the two sides, but this did not work.
 A: Write the equation as
$$
\left(\frac{1}{2}\right)^x+\left(\frac{\sqrt{3}}{2}\right)^x=1.
$$
and note that the LHS is strictly decreasing as $0<1/2<1$ and $0<\sqrt{3}/2<1$ . Hence the equation has at most one solution. It is clear that $x=2$ is a solution.
A: $$(\cos60^\circ)^x+(\sin60^\circ)^x=1$$
Now for $0<u<90^\circ,$
Can you prove
$$(\cos u)^m+(\sin u)^m$$ is decreasing function so that the solution is unique
A: put x = 2 in above expression . $1+3^{\frac{2}{2}} = 1+3^1 = 4$
$2^2 = 4 $
$x = 2 $ is a solution
A: Let $x = 2 y$ then the equation becomes
$$f(y)=4^y - 3^y = 1$$
for which only $y=1$ is the only solution. This leads to $x = 2$ as the only solution of the desired equation, since $f(y)$ will only increase as $y$ increases
A: You can also divide both sides by $(\sqrt{3})^x$.
$$\left(\frac{1}{\sqrt{3}}\right)^x+1=\left(\frac{2}{\sqrt{3}}\right)^x$$
$0<\frac{1}{\sqrt{3}}<1$ and $1<\frac{2}{\sqrt{3}}$, so the left side is strictly decreasing and the right side is strictly increasing, so the equation has at most one solution and $x=2$ is a solution.
