Question

If x is a real number such that log3 5 = log5 (2 + x), then which of the following is true?

• A. 0 < x < 3
• B. 23 < x < 30
• C. x > 30
• D. 3 < x < 23

Answer 1

The Correct Option is D

To solve the problem, we need to find the value of \( x \) given the equation \( \log_3 5 = \log_5 (2 + x) \). Let's break down the steps:

1. Rewrite the equation in a common form using the change of base formula:
   • The change of base formula for logarithms is: \(\log_a b = \frac{\log_c b}{\log_c a}\) for any positive base \( c \) not equal to 1. In this problem, we can use base 10 or any common base.
   • So, the left side becomes: \( \log_3 5 = \frac{\log_{10} 5}{\log_{10} 3} \).
   • The right side is: \( \log_5 (2 + x) = \frac{\log_{10} (2 + x)}{\log_{10} 5} \). 
2. Set the equations equal to each other:
   • \(\frac{\log_{10} 5}{\log_{10} 3} = \frac{\log_{10} (2 + x)}{\log_{10} 5}\)
3. Cross-multiply to clear the fractions:
   • \((\log_{10} 5)^2 = (\log_{10} 3) \cdot (\log_{10} (2 + x))\)
4. Take the exponent of both sides to solve for \( 2 + x \):
   • Raise the base 10 to the powers of both sides, leading to: \( 5^2 = 3 \cdot (2 + x) \)
   • This simplifies to: \( 25 = 3(2 + x) \)
5. Solve for \( x \):
   • Distribute and rearrange: \( 25 = 6 + 3x \)
   • Subtract 6 from both sides: \( 19 = 3x \)
   • Divide by 3 to isolate \( x \): \( x = \frac{19}{3} \approx 6.33 \)
6. Determine which option the solution falls into:
   • \( \frac{19}{3} \approx 6.33 \), so the correct range is \( 3 < x < 23 \).

Therefore, the correct answer is: 3 < x < 23.