Solve the equation 5^(x^2) - 5^(-5x+6) = 0

Expert Answers
hala718 eNotes educator| Certified Educator

5^x^2 - 5^(-5x+6) = 0

==> 5^x^2 = 5^(-5x + 6)

==> x^2 = -5x + 6

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

==> (x+6)(x-1) = 0

==> x1= -6

==> x2= 1

 

giorgiana1976 | Student

This is an exponential equation.

We'll move the second term to the right side. We notice that the bases from the both sides are matching, so we'll use the one to one property of exponential functions:

5^(x^2) = 5^(-5x+6) => x^2 = -5x+6

We'll move all terms to one side:

x^2 + 5x - 6 = 0

We'll apply the quadratic formula:

x1 = [-5+sqrt(25-24)]/2

x1 = (-5+1)/2

x1 = -2

x2 = (-5-1)/2

x2 = -3

We'll check the solutions into the original equation:

5^(x^2) = 5^(-5x+6) for x1 = -2

5^(4) = 5^(10+6)

5^4 = 5^16 impossible

5^(9) = 5^(15+6) for x2 = -3

5^9 = 5^21 impossible

So, both solutions are not valid!

neela | Student

5^(x^2) - 5^(-5x+6) = 0.

To solve for x.

Solution:

Rewrite the equation:

5^(x^2) = 5^(-5x+6). Since the bases are same , i.e, 5 in both sides, the exponents of 5  should be equal.

x^2 = -5x+6.

x^2+5x-6 = 0

x^2+6x-x-6 = 0

x(x+6)-1(x+1) = 0

(x+6)(x-1) = 0

 x+6 = 0 , x-1 = 0

x = -6 or x =1.

 

thewriter | Student

We have to solve 5^(x^2)-5^(-5x+6)=0

First express the equation as 5^(x^2)=5^(-5x+6)

Now take the logarithm of both the sides. We use the relation log(x^a)=a logx

Therefore we get x^2log5=(-5x+6)log5

=>x^2=(-5x+6)

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

The roots of this equation are

[-5+sqrt(25-24)]/2=(-5+1)/2=-2

and [-5-sqrt(25-24)]/2=(-5-1)/2=-3

But substituting -2 in x^2=(-5x+6) 4= 16, which is absurd.

Also substituting -3 in x^2=(-5x+6) gives 9=21, again not possible.

So the equation does not have any valid solution.