Find the solutions for x

Take $5^{x}=p$

Then, $p^{2}+5^{5}=5^{2}p+5^{3}p$ .
$(p-25)×(p-125)=0$

Then, $p=5^{2} , p=5^{3}$

Then $x=3,x=2$

Taking the original equation, I want to isolate $5^{x}$

So I rewrite it: $(5^{x})^{2} + 5^{5} = (5^{x} )(5^{3} ) + (5^{x} )(5^{2}) .$

Now I sub in y for $5^{x}$ and rearrange it a bit to make it into a simple quadratic equation: $y^{2} - (5^{3} + 5^{2} )y + (5^{3})(5^{2}) = 0$ .

Now I can easily factorise this equation:

$(y - 5^{3} ) (y - 5^{2}) = 0$

$y = 5^{3}$ OR $y = 5^{2}$

Then sub in $5^{x}$ again :

$5^{x} = 5^{3}$ $5^{x} = 5^{2}$

$x = 3$ OR $x = 2$

So the sum of the solutions is $3 + 2 = 5.$

1) if x+3=5 & x+2=2x, both are satisfied then x is a solution 2) if x+3=2x & x+2=5 both are satisfied then x is a solution, Case 1 gives x=2, and case 2 gives x=3, so the sum is 3+2=5.

All the bases are 5 and the exponents on each side add up to the same total. So we must allocate the exponents across the two terms in the same way on each side or the sums will not match, although we can change the order. So the 5^5 term either matches 5^(X+3) or 5^(X+2). The two solutions are 2 and 3. The sum is 5.

Note that the exponents on both sides add up to the same value: $2x + 5$ .

Note that the function $f(x) = 5^x$ is convex. Hence, if a + b = c + d with $a < c \leq d < b$ , we have $5^a + 5^b > 5^c + 5^d.$

Thus, to obtain equality, we must have $\{ 2x, 5\} = \{x+3, x+2\}$ . Considering the 2 possible permutations, we obtain x = 2 or 3.

This Question can also be done by applying log (base 5) both sides .

wait...

5^{2x} + 5^{5} = 5^{x + 3} + 5^{x + 2}

5^{2x + 5} = 5^{2x + 5}

x = anything

Can someone tell me where I went wrong?