If $2f(x)+f\left(\frac{1}{x}\right)=\log x$, for all $x>0$, then $f(e^x)$ is:

Given the functional equation:
$$2f(x)+f\left(\frac{1}{x}\right)=\log x\dots (1)$$
Replacing $x$ by $\frac{1}{x}$ in equation (1):
$$2f\left(\frac{1}{x}\right)+f(x)=\log \left(\frac{1}{x}\right)$$
Since $\log \left(\frac{1}{x}\right)=-\log x$:
$$2f\left(\frac{1}{x}\right)+f(x)=-\log x\dots (2)$$
Multiplying equation (1) by $2$:
$$4f(x)+2f\left(\frac{1}{x}\right)=2\log x\dots (3)$$
Subtracting equation (2) from equation (3):
$$(4f(x)-f(x))+\left(2f\left(\frac{1}{x}\right)-2f\left(\frac{1}{x}\right)\right)=2\log x-(-\log x)$$
$$3f(x)=3\log x⟹f(x)=\log x$$
Now, substitute $x=e^x$:
$$f(e^x)=\log (e^x)=x$$