How to compute the unknown power x in $0.15^x + 0.36^x = 1$ More generally, I have two known variables $a$ and $b$ and one unknown variable $x,$ where:
$$a^x + b^x = 1, a,b \in [0, 1]$$
Is there a way to compute the value of $x$ analytically?
 A: If by "analytically", you mean an exact algebraic solution, then no.
Note that $0.36^x = 0.15^{\frac{\log 0.36}{\log 0.15}\cdot x} = (0.15^x)^{\frac{\log 0.36}{\log 0.15}}$.
If we let $k = \frac{\log 0.36}{\log 0.15}$ and $0.15^x = y$ we can rewrite the original equation as
$y + y^k = 1$
For which you may be able to find nice solutions for some rational values of $k$. But this $k$ is transcendental, which precludes such a nice, exact solution.
A: Consider that we look for the zero of function
$$f(x)=a^x+b^x-1$$ As already said in answers, for the most general case, there is no analytical solution.
However, we can make very good approximations considering instead that we look for the zero of function
$$g(x)=\log(a^x+b^x)$$ which much closer to linearity than $f(x)$. We have
$$g'(x)=\frac{a^x \log (a)+b^x \log (b)}{a^x+b^x}$$
$$ g''(x)=\frac{a^x \log ^2(a)+b^x \log ^2(b)}{a^x+b^x}-\frac{\left(a^x \log (a)+b^x \log
   (b)\right)^2}{\left(a^x+b^x\right)^2}$$
What is interesting to notice is that
$$g(0)=\log(2) >0 \qquad \text{and}\qquad f''(0)=\frac{1}{4} \log ^2\left(\frac{a}{b}\right)>0$$ By Darboux theorem, this means that Newton method would converge without never overshooting the solution.
We could use Halley method and have as an estimate
$$x=-\frac{4 \log (2) \log (a b)}{(2-\log (2)) \log ^2\left(\frac{a}{b}\right)+8 \log
   (a) \log (b)}$$
Applied to the case $a=0.15$, $b=0.36$ this would give
$x=0.490244$ while the "exact" solution is $x=0.490642$.
Not exact but so close.
Doing the same with Householder method would give $x=0.490752$.
Edit
Please, notice that all the above has been obtained starting at $\color{red}{x=0}$. Assuming, without loss of generalization $a<b$, we can improve the process noticing that the solution is such that
$$\alpha=-\frac{\log (2)}{\log (a)} < x < -\frac{\log (2)}{\log (b)}=\beta$$
Taking into the quasi-lineartion of function $g(x)$, using the secant, you can generate a better estimate
$$\gamma=\frac{\beta \, g(\alpha )-\alpha \, g(\beta )}{g(\alpha )-g(\beta )}$$ and, using Halley method with $x_0=\gamma$, obtain the approximation
$$x=\gamma-\frac{2 g(\gamma )\, g'(\gamma )}{2 g'(\gamma )^2-g(\gamma )\, g''(\gamma )}$$
Applied to the working case, this would give
$$\alpha=0.365368\cdots \qquad \beta=0.678458\cdots \qquad \gamma=0.492216\cdots$$ from which the estimate
$$x=0.4906415695461$$ while the "exact" solution is
$$x=0.4906415695447$$
A: To start out with we can graph the LHS in a graphing cqalculator:

Clearly, the answer is somewhere close to $x \approx 0.5$ for $y=1$. (Specifically, close to $x = 0.49064).


As for obtaining the exact solution of $a^x + b^x = 1$, the given equation is transcendental (basically, non-algebraic or non-polynomial), so an analytic solution can only be found for some special cases. Your best bet would be to use some numerical nethod such as Newton-Raphson or Bisection to estimate the solution.
You can somewhat simplify the equation using logarithms:
$a^x + b^x = a^x + (a ^\frac{\ln b}{\ln a})^x =  (a^x) + (a^x)^{\frac{\ln b}{\ln a}} = 1$
When $\frac{\ln b}{\ln a} = \phi$ (say) has nice values, solutions can be found.
Example,
$\phi = 1 \implies a^x = 0.5 \implies x = \log_a 0.5$
$\phi = 0.5 \implies a^x = \frac{3}{3} - \frac{\sqrt{5}}{2} \implies x = \log_a (1.5 - \sqrt(5)/2)$
In this case however, this is not possible.
