How to prove using Mean value theorem $3^x-2^x >(1.5)^x, \:\forall x \geq 2$
I did using the following way:
Suffice it to prove that $$\left(\frac{1}{2}\right)^x+\left(\frac{2}{3}\right)^x<1, \:\forall x \geq 2$$ Consider the function: $$\begin{aligned} & f(x)=\left(\frac{1}{2}\right)^x+\left(\frac{2}{3}\right)^x-1 \\ & \Rightarrow f^{\prime}(x)=\left(\frac{1}{2}\right)^x \ln \left(\frac{1}{2}\right)+\left(\frac{2}{3}\right)^x \ln \left(\frac{2}{3}\right)<0 \end{aligned}$$ Thus $f$ is monotone decreasing in $[2, \infty)$. Thus $$f(x) < f(2)<0 \Rightarrow f(x)<0 \Rightarrow 3^x-2^x>\left(\frac{3}{2}\right)^x$$
Your solution is correct. Alternatively one can argue that for $x \ge 2$ $$ \left(\frac{1}{2}\right)^x+\left(\frac{2}{3}\right)^x = \frac 1 4 \left(\frac{1}{2}\right)^{x-2}+\frac 4 9 \left(\frac{2}{3}\right)^{x-2} \le \frac 1 4 + \frac 4 9 = \frac{25}{36} < 1 \, . $$