the root of $2^{x}$ = $2^{1.5}$ based on $2^{x^{\cos(x)}}\sqrt{\cos(x)}=2^{x}$ Can you obtain or is it plausible to find the roots of
if $2^{x^{\cos(x)}}\sqrt{\cos(x)}=2^{x}$
$x > 0$ & $cos(x) > 0$
what does $x$ equal in $2^{x}$ = $2^{1.5}$ exactly?
 A: Part One:
$$2^x = 2^{1.5} \Rightarrow x=1.5$$
Take the natural logarithm of both sides to see this.
Part Two:
$$f(x) = 2^{x^{\cos(x)}}\sqrt{\cos(x)} - 2^{x}$$
Because $0 < \cos(x) \leq 1$, we know that
$$2^{x^{\cos(x)}}\sqrt{\cos(x)} \leq 2^x$$
And equality is achieved where $x$ is a multiple of $2\pi$.
So your solutions are $x = 2n\pi$ for $n \geq 1$.
A: For the second half of your question, the answer numerically is obvious: $2^x=2^1.5 \rightarrow x=1.5$. This means basically taking the number and multiplying by its square root.
$$\sqrt{x}=x^{0.5} \rightarrow x\sqrt{x}=xx^{0.5}=x^{1+0.5}=x^{1.5}$$
For the first part of your question, I would recommend a numerical solution. The problem looks too hard to be tractable analytically. To find the roots, use Newton's method as shown below:
$$f(x)=2^{x^{\cos(x)}}\sqrt{\cos(x)}-2^x$$
Newton's method solves for $f(x)=0$ via the following iterative method with some initial guess $x_0$.
$$x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$$
You can analytically differentiate the expression you had above, if you apply chain rule multiple times. If you are not familiar with derivatives, you can use a simple numerical approximation to $f'(x)=\frac{f(x+h)-f(x)}{h}$ for some small value of $h \approx 0.0001$. Then, the formula becomes:
$$x_{i+1}=x_i-\frac{f(x_i)h}{f(x_i+h)-f(x_i)}$$
This expression you could plug into your calculator and produce a solution to pretty much whatever degree of accuracy you desire.
