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)

Asked on by edobro

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Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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).

oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

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!

 

edobro's profile pic

edobro | Student, Grade 11 | (Level 2) Honors

Posted on

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.

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