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.
On
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.
On
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!
Hint: Analyse this function and try to prove $x=2$ is only solution $$f\left( x \right) =4\cdot 2^{ x }+3^{ x }-5^{ x }$$