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

To solve the given equation, \(\log_5(x-2) = 2\log_{25}(2x-4)\), we need to analyze and simplify both sides to determine the values of \( x \) for which the equation holds.

1. On the right side of the equation, note that \(\log_{25}(2x-4)\) can be rewritten using the change of base formula:

\(\log_{25}(2x-4) = \frac{\log_5(2x-4)}{\log_5(25)}\)

1. Since \(25 = 5^2\), we have \(\log_5(25) = 2\). Therefore:

\(\log_{25}(2x-4) = \frac{\log_5(2x-4)}{2}\)

1. Substitute back into the equation:

\(\log_5(x-2) = 2 \times \frac{\log_5(2x-4)}{2}\)

1. Simplifying gives:

\(\log_5(x-2) = \log_5(2x-4)\)

1. Since the logarithms have the same base and are equal, this implies:

\(x-2 = 2x-4\)

1. Simplifying the equation:

\(-x = -2\)

1. Therefore:

\(x = 2\)

1. However, it's important to check the domain of the original logarithmic expressions:
   • For \(\log_5(x-2)\) to be defined, \(x-2 > 0 \rightarrow x > 2\)
   • For \(\log_{25}(2x-4)\) to be defined, \(2x-4 > 0 \rightarrow x > 2\)
2. The value \(x = 2\) does not satisfy the condition \(x > 2\). Therefore, there are no real values of \( x \) that satisfy the original equation.

Thus, the correct answer is that there are 0 real solutions for the given equation.

Answer 2

To solve the given equation, \( \log_5(x-2) = 2\log_{25}(2x-4) \), let's break it down step by step to understand how many solutions exist for the variable \( x \).

1. First, rewrite the equation using the change of base formula and properties of logarithms:
   • The logarithm on the left-hand side is already in base 5. We need to express the right-hand side in terms of base 5 as well.
   • The right-hand side is \( 2\log_{25}(2x-4) \). Notice that \( 25 = 5^2 \), so we can write: \(\log_{25}(2x-4) = \frac{1}{2}\log_5(2x-4)\)
   • Thus, the equation becomes: \(\log_5(x-2) = 2\left(\frac{1}{2}\log_5(2x-4)\right)\)
   • This simplifies to: \(\log_5(x-2) = \log_5(2x-4)\)
2. Since the equation \(\log_5(x-2) = \log_5(2x-4)\) implies that the arguments must be equal (because the logarithmic function is one-to-one), we have: \(x-2 = 2x-4\)
3. Solving the equation \( x-2 = 2x-4 \):
   • Rearrange the terms: \(x - 2x = -4 + 2\)
   • This simplifies to: \(-x = -2\)
   • Multiply both sides by -1: \(x = 2\)
4. Now, substitute \( x = 2 \) back into the original equation to check for possible values:
   • At \( x = 2 \), the left-hand side becomes: \(\log_5(2-2) = \log_5(0)\), which is undefined.
   • Therefore, \( x = 2 \) is not a valid solution because the logarithm of zero is undefined.

Since substituting \( x = 2 \) makes the original equation undefined and no other \( x \) will satisfy the given mathematical equality under the real number system, there is 0 value of \( x \) that satisfies the equation.