How to find $x$ when $2^{x}+3^{x}=6$? $$2^{x}+3^{x}=6$$
How to find the real number x? I mean it's approximately $1.19$ bur can we write $x$ as the form of $a, b, c$ when $a^{x}+b^{x}=c$ in general. Maybe an infinite sum?
 A: This kind of equation does not show analytical solution and only numerical methods would be able to solve the problem.
What is nice if that we know the range which contains the bounds $$1 < x < \frac{\log (3)}{\log (2)}$$
However, since the function is quite stiff (it varied very fast), it should be better to consider the function $$f(x)=\log(2^x+3^x)-\log(6)$$ which looks like a straight line. So, let me be lazy and start iterating from $x_0=1$. Then Newton iterates will be $$x_1=1.19469932382260$$ $$x_2=1.19391148996091$$ $$x_3=1.19391147721159$$ which is the solution for fifteen significant figures.
For the more general case of $a^x+b^x=c$, assuming $a>0,b>0,c>0$, we know that the solution will be such that $$\min \left(\frac{\log \left(\frac{c}{2}\right)}{\log (a)},\frac{\log
   \left(\frac{c}{2}\right)}{\log (b)}\right)<x<\max \left(\frac{\log \left(\frac{c}{2}\right)}{\log (a)},\frac{\log
   \left(\frac{c}{2}\right)}{\log (b)}\right)$$ and, starting at the middle of the range, we can start Newton procedure for $$f(x)=\log(a^x+b^x)-\log(c)$$ If we can start using $x_0=1$, the first iterate of Newton method will be $$x_1=\frac{-(a+b) (\log (a+b)-\log (c))+a \log (a)+b \log (b)}{a \log (a)+b \log (b)}$$
A: There is a way to build an (approximate) explicit formula, using Euler's method of substitution (I think Euler was the first to use it anyway).
First, we need to transform our equation. Let's take $3^x=y$ and $\alpha=\frac{log(2)}{log(3)}$ with $log$ being the natural logarithm. Then we have:
$$
y^\alpha+y-6=0
$$
$$
y=6-y^\alpha
$$
$$
y^\alpha=(6-y^\alpha)^\alpha
$$
Substitute $y^\alpha$ into the right side several times:
$$
y \approx 6-\left(6-\left(6-\left(6-\left(6-... \right)^\alpha \right)^\alpha \right)^\alpha \right)^\alpha
$$
Then we calculate $\alpha=0.63092975357$ and calculate $y$.
This expression is very inefficient - it converges very slowly, but it will get you the approximate value easily enough. The accurate solution is:
$$
y=3^x=3^{1.1939115}=3.71228
$$
The first few values given by our expression are:
$$
y_1=6-6^\alpha=2.903...
$$
$$
y_2=6-(6-6^\alpha)^a=4.041...
$$
$$
y_3=3.586...
$$
$$
y_4=3.762...
$$
$$
y_5=3.693...
$$
$$
y_6=3.720...
$$
$$
y_7=3.709...
$$
As you can see, it's pretty bad. But seeing as odd values are smaller than the accurate solution and even values are larger, you can use the mean value for better approximation.
Anyway, you wanted the general expression for $a^x+b^x=c$ - here is the way you can make it.
Please note, that for convergence the following condition should hold: either $c>1$ and $\alpha<1$ or $c<1$ and $\alpha>1$. For any other cases, the expression will not work.
