Base five nuneration system

Let $5^{x} =y$

. So given eqn becomes $5y + \frac{5}{y} = 26$

$\implies 5.y^{2} - 26 y + 5 =0$

$(y-5)(y-\frac{1}{5})=0$ .

$\therefore y=5 \ or \ y=\frac{1}{5}$

so $x=1 \ or \ x=-1$

Hence $+1-1 = \boxed{0}$ .

enjoy!

Note that the equation is symmetric in $x$ and $-x$ . Hence, if $x=a$ is a solution, then so is $x=-a$ . So, the sum of all solutions is zero.

Steps for this question.

Steps for this question (just by plugging in).

Let´s Think about it...

$5^{(x+1)} + 5^{(1-x)} = 26 = 25 + 1 = 5^2+5^0$

$\implies$ { $(x+1) = 2$ and $(1-x) = 0$ } or { $(x+1) = 0$ and $(1-x) = 2$ }

$\implies$ { $x = 1$ and $1 = x$ } or { $x = -1$ and $-1 = x$ }

So, $x=\{-1,1\}$

Considering the result is ending with 6 and any product of 5 will end with a 5. Therefore a 1 is needed in the addition so one of the exponant has to be 0.

I have just put the values (1,-1) and checked the equation. When we add them they become Zero (0).

Just split 26 as 5^2+5^0

Equate (x+1)=2

Then equate (x+1)=0

Since there are the two solution,we can conclude that the solutions 1 and -1 respectively