Question

For a real number \(x>1\), solve the equation: \[ \frac{1}{\log_2x} +\frac{1}{\log_3x} + \frac{1}{\log_4x} = 1 \] The value of x is:

• A. 4
• B. 12
• C. 24
• D. 36

Hint:

To simplify logarithmic equations, apply the change of base rule and logarithmic identities effectively.

Answer 1

The Correct Option is C

Step 1: Apply the change of base formula to rewrite the logarithmic expressions.
Using the formula \(\log_a x = \frac{\log x}{\log a}\), the equation is transformed into: \[ \frac{\log 2}{\log x} + \frac{\log 3}{\log x} + \frac{\log 4}{\log x} = 1 \] Step 2: Simplify the given equation.
Factor out \(\frac{1}{\log x}\) from the terms: \[ \frac{1}{\log x} \left( \log 2 + \log 3 + \log 4 \right) = 1 \] Combine logarithmic terms using the property \(\log a + \log b = \log(ab)\): \[ \log 2 + \log 3 + \log 4 = \log(2 \times 3 \times 4) = \log 24 \] Thus, the equation simplifies to: \[ \frac{\log 24}{\log x} = 1 \] Step 3: Determine the value of \(x\).
Multiplying both sides by \(\log x\) gives: \[ \log 24 = \log x \] Applying the logarithmic property where \(\log a = \log b\) implies \(a = b\), we get: \[ x = 24 \] Final Answer: \[\boxed{{(C) 24}}\]