How to find the solution to $4^x + 9^x = 4?$ I tried to solve this equation algebraically but couldn't. Maybe there isn't a way to extract the solution from the equation algebraically. If it's the latter, I'd really like to know why. Also, is there a general solution to equations with the form $a^x + b^x = c$ where $a, b,$ and $c$ are all different real numbers? Thanks.
 A: Here's a quick way to get bounds on the solution in the case $b \ge a > 1$, $a + b > c > 1$. Since we clearly must have $0 < x < 1$, we can use the power mean inequalities to get
$$
\sqrt{ab} \le \left(\frac{a^x+b^x}{2}\right)^{1/x}\le \frac{a+b}{2}
$$
Substituting $c$ in for $a^x +b^x$, taking natural logs, and solving for $x$ gives
$$
 \frac{\ln (c/2)}{\ln[(a+b)/2]}\le x\le \frac{2\ln (c/2)}{\ln(ab)}.
$$
Applying this to your case of $a = c = 4$ and $b = 9$ gives $0.370\le x \le 0.387$. Taking the midpoint of this interval gives $0.3786 \pm 0.0083$, an estimate with a $2.2\%$ margin of error. The actual error from the true value $0.38024$ is about $0.4\%$.
This method works best if $a$ is not close to $1$ and $b$ is not much larger than $a$, but the bounds it gives will always be correct. In particular, the percent error is bounded above by $\ln\left(\frac{a+b}{2\sqrt{a b}}\right)/\ln(a b)$, though showing that is a bit more involved.
Lastly, if you really wanted more accuracy, you could actually perform this method iteratively. That is, since now you know $p \le x \le q$, you can use the power mean inequality again to get
$$
\frac{q\ln(c/2)}{\ln[(a^q+b^q)/2]}  \le x \le \frac{p\ln(c/2)}{\ln[(a^p+b^p)/2]}
$$
These bounds converge exponentially, making it a reasonably efficient algorithm for computing the value of $x$.
A: This is not a proper solution to the problem
Draw the graph of $4^x+9^x$ and $y=4$
Let the co-ordinates of the point where they intersect be $(a,4)$ Now $a$ is the required answer. By estimating roughly and taking some help, the value of $a$ is expected to come around $$0.38$$
Some additional facts that may be useful $$4^x+9^x-4=-2+\sum_{n=1}^{\infty}\frac{(x^n)(\operatorname{log}^n(4)+\operatorname{log}^n(9))}{n!}$$
which is also equal to
$$9+\sum_{n=1}^{\infty}\frac{(-1+x)^n(4\operatorname{log}^n(4)+9\operatorname{log}^n(9))}{n!}$$

This is the complex map of the equation.
The Graph $($credit to PM $2$ Ring$)$

A: Instead of looking at the equation, consider that you look for the zero of function
$$f(x)=4^x+9^x-4$$
The solution is bounded by
$$4^x+4^x=4 \implies x_1=\frac 12\implies f(x_1)=1$$
and by
$$9^x+9^x=4\implies x_2=\frac{\log (2)}{\log (9)}\implies f(x_2)=4^{\frac{\log (2)}{\log (9)}}-2=-0.451437$$
If you plot it, you will notice that $f(x)$ is very linear for $x \in (x_2,x_1)$. Then, draw a straight line joining these two points : this will give an $x$ intercept close to $x_3=0.372860$ (notice that $f(x_3)=-0.054372$ - big improvement if you think that the solution is $x=0.380246$).
Repeat the process with points $(x_2,x_3)$; now $x_4=0.380720$.
Nothing very complicated. Very soon, you will learn how we can do it much faster. Just to give you a taste of it, using Newton method starting with $x_2$, the iterates will be
$$\{0.31546488,0.38447920,0.38026302,0.38024568\}$$
