Solving for the exponent of a power sum Let $x$, $y$, $z$, $t$, be real positive numbers.
Is it possible to solve $t$ from the following equation, if $x$, $y$, and $z$ are known?
$$
x^t + y^t=z
$$
If an exact solution is not possible, are there good approximations?
The more general case of interest would be to find $t$ that satisfies
$$
\sum_i a_i^t b_i=c
$$
Assuming that $a_i$, $b_i$ and $c$ are all known. We also know that $\sum_i b_i=1$ and that $a_i>1$.
 A: You can attempt to find the answer through the Fixed-Point method.
$$x^t+y^t=z$$
$$x^t=z-y^t$$
$$t=\log_x(z-y^t)$$
From here, you can use $t_{n+1}=\log_x(z-y^{t_n})$.
Assume you know what $t$ is, approximately.  Use that as $t_0$.  You will note that $t_n$ will either converge or diverge from the correct answer for $t$.
If it diverges, try $t_{n+1}=\log_y(z-x^{t_n})$.  It might work, though it really depends on what the correct value of $t$ is and what $x,y,z$ are.
Also, if you have to deal with complex numbers, I will tell you that it may converge, even if there are complex numbers.
For the general case:
$$t_{n+1}=\log_{a_k}(\frac{c-\sum_ia_i^tb_i-a_k^tb_k}{b_k})$$
Choosing $k$ to be a value that makes the series converge.
More ideas
If we use the substitution $t=\log(u)$, we get
$$x^{\log(u)}+y^{\log(u)}=z$$
$$u^{\log(x)}+u^{\log(y)}=z$$
We can use this adjustment to try and solve for $u$, given $x,y$, we can make the base of the logarithm something that could be simplified.
We could also attempt the Fixed-Point method again:
$$u_{n+1}=\sqrt[\log(x)]{z-u_n^{\log(y)}}$$
If we let $\log(x)=x'$ and $\log(y)=y'$, then it is equivalent to solving a polynomial of the form $u^{x'}+u^{y'}=z$, which cannot be done for general and most certainly cannot be done for $x',y'$ that are not whole numbers.  Though, sometimes you may be able to simplify things.
Also, $u_{n+1}=\sqrt[x']{z-u_n^{y'}}$
