# solve this exponetial equation: (1/25)^(-x^2+5x-8)<(1/25)^-2 This inequation isnt supposed to be solved by log or ln.Note the expressions 1.25 is raised to the power of all this expression...

solve this exponetial equation: (1/25)^(-x^2+5x-8)<(1/25)^-2

This inequation isnt supposed to be solved by log or ln.

Note the expressions 1.25 is raised to the power of all this expression -x^2+5x-8

The answer for me was (2,3) but my friends result was (3,+infinity) so when i used to prove his solution it seemed to be wrong.Anyone who is able to solve this please help, i want wheather im right or wrong.

P.s my friend used the principle rule. a^f(x)>a^g(x)

if a<1 than f(x)<g(x)

*print*Print*list*Cite

Yes, the principle regarding exponential function, used by your friend, is helpful.

Let's see how this principle is used. Since the base 1/25 is smaller than 1, the exponential function is decreasing, therefore if (1/25)^(-x^2+5x-8)<(1/25)^-2 => -x^2+5x-8 > -2.

We'll solve the inequation -x^2+5x-8 > -2

We'll move all the terms to the left side:

-x^2 + 5x - 8 + 2 > 0

-x^2 + 5x - 6 > 0

We'll multiply by -1:

x^2 - 5x + 6 < 0

We'll determine the zeroes of the equation x^2 - 5x + 6 = 0.

x1 = 3 and x2 = 2

**Since the parabola is going under x axis over the interval (2;3), then the inequality is verified if x belongs to the interval (2;3).**

indeed:

`(1/25)^(-x^2+5x-8)< (1/25)^(-2)` (1)

Your friend, according me, did trick himself.

Indeed, even if (1) implies:

`-2< -x^2+5x-8` for `(1/25)<1`

that is: `-x^2+5x-6>0`

On the other side you can write (1) as:

`1/25^(-x^2+5x-8)< 1/25^-2`

`1/25^(-(x^2-5x+8)) < 1/25^(-2)`

`25^(-(-(x^2-5x+8)))<25^(-(-2))`

`25^(x^2-5x+8)<25^2`

`x^2-5x+8<2`

`x^2-5x+6<0`

that is :

`-x^2+5x-6>0`

So: WHATEVER WAY YOU CHOICE THE RESLULT IS TO BE THE SAME!

Yes this is excatly how i used to solve it but the strange result of my and the professor said it was right how he solved that inequation, but thank you very much for helping me prove the right answer.