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 equation. The given equation is:

log₃(<i>x</i> − 2) = 2log₂₅(2<i>x</i> − 4)

Apply the property a log_b(c) = log_b(c^a) to rewrite the equation as:

log₃(<i>x</i> − 2) = log₂₅((2<i>x</i> − 4)²)

Step 2: Unify the logarithmic bases. Using the change of base formula log_b(a) = $\frac{\log(a)}{\log(b)}$, the equation transforms to:

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

Further simplification yields:

log(<i>x</i> − 2) ⋅ log(25) = log(3) ⋅ log((2<i>x</i> − 4)²)

Step 3: Solve for x. Expand log((2<i>x</i> − 4)²) using the property log(a^b) = b log(a):

log(<i>x</i> − 2) ⋅ log(25) = 2log(3) ⋅ log(2<i>x</i> − 4)

Let u = log(x − 2) and v = log(2x − 4). Substitute these into the equation:

u ⋅ log(25) = 2v ⋅ log(3)

The solutions obtained are x = 3 and x = 5. Thus, there are 2 solutions to the equation.

Answer: 2