By making the substitution u =ln x show that the following equation has exactly one real-number solution:
9/ln x + ln x^4 =-12
9/u + 4u = -12
ln ( x ^ 4) = 4 ln x
9 / ln x + ln ( x ^ 4 ) = - 12
9 / u + 4 u = - 12 Multiply both sides by u
9 u / u + 4 u * u = ( - 12 ) * u
9 + 4 u ^ 2 = - 12 u Add 12 u to both sides
9 + 4 u ^ 2 + 12 u = - 12 u + 12 u
4 u ^ 2 + 12 u + 9 = 0
Now, find discriminant using formula D = b ^ 2 − 4 a c
In this case: a = 4, b = 12 , c = 9
D = 12 ^ 2 - 4 * 4 * 9 = 144 - 144 = 0
Since discriminant is zero, then there is one root.
To find it, use formula u = - b / 2 a = - 12 / 2 * 4 = - 12 / 8 = - 4 * 3 / 4 * 2 = - 3 / 2
u = ln x
so
ln x = - 3 / 2
To show that the equation has exactly one real-number solution, we will apply the substitution u = ln x.
Step 1: Substitute u = ln x in the given equation:
9/u + 4u = -12
Step 2: Multiply both sides of the equation by u:
9 + 4u^2 = -12u
Step 3: Rearrange the equation and set it equal to zero:
4u^2 + 12u + 9 = 0
Now, we have a quadratic equation in terms of u.
Step 4: Solve the quadratic equation. Applying the quadratic formula, we get:
u = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 4, b = 12, and c = 9.
u = (-12 ± √(12^2 - 4 * 4 * 9)) / (2 * 4)
u = (-12 ± √(144 - 144)) / 8
u = (-12 ± √0) / 8
Since the discriminant (b^2 - 4ac) is zero, the quadratic equation only has one real-number solution.
Therefore, the equation has exactly one real-number solution when u = ln x.