By making the substitution u =ln x show that the following equation has exactly one real-number solution:

-12/ln x +ln x^4 = -9

well, do the math:

-12/u + 4u = 9
4u^2-9u-12=0

solve for u and then consider the domain of ln x.

I don't understand how you got those equations! Pleaseee help

To show that the equation -12/ln x + ln x^4 = -9 has exactly one real-number solution, we can start by making the substitution u = ln x.

Substituting u = ln x into the equation, we get:
-12/u + 4u = -9

Now, let's manipulate the equation to solve for u. Multiply both sides by u to eliminate the fraction:
-12 + 4u^2 = -9u

Rearrange the equation to form a quadratic equation:
4u^2 + 9u - 12 = 0

To find the solutions for u, we can use the quadratic formula:
u = (-b ± sqrt(b^2 - 4ac)) / (2a)

For our equation, a = 4, b = 9, and c = -12. Substituting these values into the formula:
u = (-9 ± sqrt(9^2 - 4*4*(-12))) / (2*4)
u = (-9 ± sqrt(81 + 192)) / 8
u = (-9 ± sqrt(273)) / 8

Since the discriminant (b^2 - 4ac) is positive and the coefficients are real numbers, the square root of 273 is a real number. Therefore, there are two possible solutions for u.

However, we made the substitution u = ln x. To find the corresponding values of x, we need to reverse the substitution.

For the first solution:
ln x = u = (-9 + sqrt(273)) / 8

Using the property of logarithms, we can solve for x:
x = e^ln x = e^((-9 + sqrt(273)) / 8)

For the second solution:
ln x = u = (-9 - sqrt(273)) / 8

Once again using the property of logarithms:
x = e^ln x = e^((-9 - sqrt(273)) / 8)

So, the original equation -12/ln x + ln x^4 = -9 has exactly one real-number solution since the substitution u = ln x leads to two different values for x.