Solve $4 \times2^x+3^x=5^x$ without any sort of calculator Is there a way i can solve the following equation only by using high school mathematics?
$$4 \times2^x+3^x=5^x$$
I tried writing $5$ as $2+3$ but didn't get any result.
After that i tried to divide by $5^x$ and see how the function goes, but, unfortunately didn't got me somewhere either.
 A: Assuming $x$ to be natural,
$$
5^x-3^x=(5-3)(5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^x\\
\implies (5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^{x-1}
$$
Each term in LHS looks like $3^{n}5^{x-1-n}>2^n2^{x-1-n}=2^{x-1}$.
There are $x$ number of terms in LHS. If there are $4$ or more terms in LHS, each of them would be greater than $2^{x-1}$ and the equation would not hold. So $1\le x\le 3$. Manually checking $x=1,2,3$ we see $x=2$ is a solution.
A: Hint: Divide both sides by $5^x$, and notice that the left side is strictly decreasing, since it is the sum of two strictly decreasing functions, and the right side is constant. It follows that our equation has at most one solution. Can you find it ? ;-$)$
A: Hint: Analyse this  function and try to prove  $x=2$ is only solution  $$f\left( x \right) =4\cdot 2^{ x }+3^{ x }-5^{ x }$$ 
A: You tried to use $2+3=5$ to no avail. A slightly different "flash of insight" approach would be to invoke Pythagoras: $3^2+4^2=5^2$.
A: At first I thought it was rather simple if you use the logarithm rules. I must have made a mistake somewhere since I am getting x = -1. Maybe my method will be helpful though so I'm posting it here:
First step: take the 3^x to RHS;
Second step: use logarithms on both sides;
Third step: rewrite 4 as 2^2 (hence your LHS is log ( 2^(2+x))   );
Fourth step: use logarithm rule on RHS thus obtain (2+x)*log(2)
;
Fifht step: use logarithm rules on LHS : log(5^x - 3^x) = log(5^x)/log(3^x) = (x * log5)/ (x * log3). The x of course cancels hence we have log5/log3;
Sixth step: divide both sides by log(2) hence your RHS = 2+x and your LHS = log5 / (log2 * log3). yet again apply log rule to the LHS hence we have log5/log5 which is 1 but then we end up with x+2 = 1 hence the x = -1. 
This is obviously not the case since it can easily be shown by plugging in x = -1 that it's not the equation. 
Good luck!
