Solve log base (x-4) of (x^2-16) = log base (x-4) of (x+2) + 2. Find the sum of all valid values of x.
Answer and solution
Correct answer: A
Solution
Answer: A — (3 + √57)/2
1. A logarithm needs a positive base other than 1 and positive arguments. Here x>4 and x≠5 ensure all these conditions.
2. Combine logarithms: x²−16=(x+2)(x−4)². Factor the left as (x−4)(x+4).
3. Since x−4≠0, divide by it: x+4=(x+2)(x−4)=x²−2x−8.
4. Solve x²−3x−12=0: x=(3±√57)/2.
5. Only (3+√57)/2≈5.275 satisfies the domain; it is not 5. Sum of valid roots=(3+√57)/2, option A.
Why this method works:
The base x−4 must be positive and unequal to 1. This gives x>4 with x≠5; both log arguments are then positive. Replace the constant 2 by log base(x−4) of (x−4)² before combining the right side into a single logarithm.
Independent check and common trap:
Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.
Apply: Write the domain first: x>4, x≠5. Cancel x−4 only after this check. Reject the negative quadratic root; the sum is the one surviving root.
Check: Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.
1. A logarithm needs a positive base other than 1 and positive arguments. Here x>4 and x≠5 ensure all these conditions.
2. Combine logarithms: x²−16=(x+2)(x−4)². Factor the left as (x−4)(x+4).
3. Since x−4≠0, divide by it: x+4=(x+2)(x−4)=x²−2x−8.
4. Solve x²−3x−12=0: x=(3±√57)/2.
5. Only (3+√57)/2≈5.275 satisfies the domain; it is not 5. Sum of valid roots=(3+√57)/2, option A.
Why this method works:
The base x−4 must be positive and unequal to 1. This gives x>4 with x≠5; both log arguments are then positive. Replace the constant 2 by log base(x−4) of (x−4)² before combining the right side into a single logarithm.
Independent check and common trap:
Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.
G Strategy shortcut
Spot: Logarithms.Apply: Write the domain first: x>4, x≠5. Cancel x−4 only after this check. Reject the negative quadratic root; the sum is the one surviving root.
Check: Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.
G Strategy: Spot: Logarithms. Apply: Write the domain first: x>4, x≠5. Cancel x−4 only after this check. Reject the negative quadratic root; the sum is the one surviving root. Check: Equal logarithms with the same valid base have equal arguments, even if the base lies between 0 and 1. Division by x−4 is legal under the domain. Of the quadratic roots, only about 5.275 is allowed; do not use the Vieta sum 3 because that includes the rejected root.