Apparently cannot be solved using logarithms

I have solved a question similar to this before. In general, you can have a solution of the equation

$$ a^x=bx+c $$

in terms of the Lambert W-function

$$ -\frac{1}{\ln(a)}W_k \left( -\frac{1}{b}\ln(a) {{\rm e}^{-{\frac {c\ln(a) }{b}}}} \right)-{\frac {c}{b}} \,.$$

Substituting $ a=1.01 \,,b=\frac{1}{2}\,,c=\frac{3}{2}$ and considering the values $k=0$ and $k=-1$, we get the zeroes $$x_1= -1.020199952\,, x_2=568.2993002 \,. $$

Polynomials don't play nice with exponentials, so no. If you work hard, you might find an answer in terms of the Lambert W function, but if I did I wouldn't feel much more enlightened.