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?
You could, for example, write the solution to $2^x + 3^x = y$ as a series in powers of $y-2$:
$$\eqalign{x &= \frac{y-2}{\ln(6)} - \left(\ln(3)^2 + \ln(2)^2\right) \frac{(y-2)^2}{2 \ln(6)^3}\cr +& \left(2 \ln(3)^4 - \ln(3)^3\ln(2)+6\ln(3)^2\ln(2)^2-\ln(3)\ln(2)^3 + 2 \ln(2)^4\right)\frac{(y-2)^3}{6 \ln(6)^5}\cr - &\left(3 \ln(3)^6 - 4 \ln(3)^5 \ln(2) + 18 \ln(3)^4 \ln(2)^2 - 10 \ln(3)^3 \ln(2)^3 + 18 \ln(3)^2 \ln(2)^4 - 4 \ln(3)\ln(2)^5 + 3 \ln(2)^6 \right) \frac{(y-2)^4}{12 \ln(6)^7} \cr + &\ldots }$$
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 $$x=\frac{y-2}{\log (ab)}-\frac{(y-2)^2 \left(\log ^2(a)+\log ^2(b)\right)}{2 \log^3 (ab)}+\frac{(y-2)^3 \left(-\log ^3(a) \log (b)-\log (a) \log ^3(b)+6 \log ^2(a) \log ^2(b)+2 \log ^4(a)+2 \log ^4(b)\right)}{6 \log^5 (ab)}+O\left((y-2)^4\right)$$
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 $$ g(x) = \ln \left( y \right) = \ln \left( {2^{\,x} + 3^{\,x} } \right)\quad g(x)' = {{\ln \left( 2 \right)2^{\,x} + \ln \left( 3 \right)3^{\,x} } \over {g(x)}} $$
Note that only bijective functions have an inverse function. Over the reals, your function is injective and we assume that it is defined surjective, hence bijective.
Functions (and inverse functions) can be represented numerically, by infinite series (inverse functions e.g. through Lagrange inversion) or by means of algebra.
My answer is for finding the inverse by the rearrangement of equations by means of algebra and in closed form. The default answer in this case is that the inverse function of this function isn't an elementary function. Note that my answer isn't the usual answer.
Your function with $$f(x)=2^x+3^x$$ is an elementary function.
1.)
We see, the function is an algebraic function in dependence of more than one algebraically independent monomials simultaneously ($2^x,3^x$). This kind of function terms is not in a form to read if the function has an inverse that is an elementary function.
Applying the formulas for definition of the inverse function, we get the following equation.
$$2^x+3^x=y$$
We see, this equation is a polynomial equation of more than one algebraically independent monomials ($2^x,3^x$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (means elementary operations) we can read from the equation. Therefore, we don't know if the function $f$ has an elementary inverse.
2.)
$$2^x+3^x=y$$ $$e^{\ln(2)x}+e^{\ln(3)x}=y$$ $x\to\ln(t)$: $$t^{\ln(2)}+t^{\ln(3)}=y$$ $$t^{\ln(2)}+t^{\ln(3)}-y=0$$
We see, this equation is a trinomial equation with real exponents. We can transform it to the form of equation 8.1 of [Belkic 2019]. Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $\ _1\Psi_1$ can be obtained therefore.
see also: Proof that there is no closed form solution to $2^x + 3^x = 10$
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