Question

Consider the equation, \(log_5(x-2) = 2log_{25}(2x-4)\) where x is a real number.
For how many different values of x does the given equation hold?

• A. 0
• B. 1
• C. 2
• D. 4
• E. Infinitely many

Answer 1

The Correct Option is A

Step 1: Simplify the given equation. The equation is:

log₃(x − 2) = 2log₂₅(2x − 4)

Using the property a log_b(c) = log_b(c^a), rewrite the equation:

log₃(x − 2) = log₂₅(2x − 4)²

Step 2: Convert the logarithmic bases to the same base. Using log_b(a) = $\frac{\log(a)}{\log(b)}$, the equation becomes:

$\frac{\log(x - 2)}{\log(3)} = \frac{\log((2x - 4)^2)}{\log(25)}$

Simplify further:

log(x − 2) ⋅ log(25) = log(3) ⋅ log((2x − 4)²)

Step 3: Solve for x. Expand log((2x − 4)²) using log(a^b) = b log(a):

log(x − 2) ⋅ log(25) = 2log(3) ⋅ log(2x − 4)

Let u = log(x − 2) and v = log(2x − 4). Substitute:

u ⋅ log(25) = 2v ⋅ log(3)

After solving, x = 3 and x = 5. Therefore, the equation has 2 solutions.

Answer: 2