Inverse of a non-trivial exponential function I am asked to determine the inverse function of this function,
    $$f(x)=2^{x}+3^{x}$$
The inverse function can not be found explicitly, since there is no way to explicitly clear x, but this does not mean that it has no inverse.
I could show a way to find the inverse of this function
Edit  How can we find the inverse of this function with the problems that it presents in the clearance of the x?
 A: You could, for example, write the solution to $2^x + 3^x = y$ as a series in powers of $y-2$:
$$\eqalign{x &= \frac{y-2}{\ln(6)} - \left(\ln(3)^2 + \ln(2)^2\right) \frac{(y-2)^2}{2 \ln(6)^3}\cr +& \left(2 \ln(3)^4 - \ln(3)^3\ln(2)+6\ln(3)^2\ln(2)^2-\ln(3)\ln(2)^3 + 2 \ln(2)^4\right)\frac{(y-2)^3}{6 \ln(6)^5}\cr - &\left(3 \ln(3)^6 - 4 \ln(3)^5 \ln(2) + 18 \ln(3)^4 \ln(2)^2 - 10 \ln(3)^3 \ln(2)^3 + 18 \ln(3)^2 \ln(2)^4 - 4 \ln(3)\ln(2)^5 + 3 \ln(2)^6 \right) \frac{(y-2)^4}{12 \ln(6)^7}
\cr + &\ldots }$$
A: This is basically the same as Robert Israel's answer.
Consider the more general case of $y=a^x+b^x$ and expand as a Taylor series around $x=0$ to get
$$y=2+x (\log (a)+\log (b))+\frac{1}{2} x^2 \left(\log ^2(a)+\log
   ^2(b)\right)+\frac{1}{6} x^3 \left(\log ^3(a)+\log ^3(b)\right)+O\left(x^4\right)$$ Now, proceed as in the link I gave in a comment to get
$$x=\frac{y-2}{\log (ab)}-\frac{(y-2)^2 \left(\log ^2(a)+\log ^2(b)\right)}{2
   \log^3 (ab)}+\frac{(y-2)^3 \left(-\log ^3(a) \log (b)-\log (a) \log
   ^3(b)+6 \log ^2(a) \log ^2(b)+2 \log ^4(a)+2 \log ^4(b)\right)}{6 \log^5 (ab)}+O\left((y-2)^4\right)$$
A: Given that
$$
y(x) = 2^{\,x}  + 3^{\,x} \quad 0 < y'(x) = \ln \left( 2 \right)2^{\,x}  + \ln \left( 3 \right)3^{\,x} 
$$
and the function is convex, in alternative to espress $x(y)$ through an infinite series,
you may consider to express it as a sequence, using e.g. the Newton-Rapson method,
i.e.
$$
x(y) = \mathop {\lim }\limits_{n\; \to \;\infty } x_{\,n}  = x_{\,n - 1}  + {{y - f\left( {x_{\,n - 1} } \right)} \over {f'\left( {x_{\,n - 1} } \right)}}\quad \left| {\;x_{\,0}  = 0} \right.
$$
which is convenient specially from the computational point of view.
You can much improve the convergency rate if you apply the method to
$$
g(x) = \ln \left( y \right) = \ln \left( {2^{\,x}  + 3^{\,x} } \right)\quad g(x)' = {{\ln \left( 2 \right)2^{\,x}  + \ln \left( 3 \right)3^{\,x} } \over {g(x)}}
$$
