Question

If \(log_4m+log_4n=log_2(m+n)\) where m and n are positive real numbers, then which of the following must be true?

• A. \(\frac{1}{m} +\frac{1}{n} =1\)
• B. \(m=n\)
• C. \(m^2+n^2=1\)
• D. \(\frac{1}{m} +\frac{1}{n} =2\)
• E. No values of m and n can satisfy the given equation

Answer 1

Answer: (E)

Step 1: Use logarithm property 

\(\log_{4} m + \log_{4} n = \log_{4}(mn)\).

Step 2: Rewrite equation

\(\log_{4}(mn) = \log_{2}(m+n)\).

Step 3: Convert bases

\(\log_{4}(mn) = \frac{\log_{2}(mn)}{\log_{2} 4} = \frac{1}{2}\log_{2}(mn)\).

So equation becomes: \[ \frac{1}{2}\log_{2}(mn) = \log_{2}(m+n). \]

Step 4: Eliminate logs

Multiply both sides by 2: \[ \log_{2}(mn) = 2\log_{2}(m+n). \]

This simplifies to: \[ \log_{2}(mn) = \log_{2}\big((m+n)^2\big). \]

Step 5: Compare arguments

\[ mn = (m+n)^2. \]

Step 6: Expand and analyze

\((m+n)^2 = m^2 + 2mn + n^2\). So: \[ mn = m^2 + 2mn + n^2. \] \[ 0 = m^2 + n^2 + mn. \]

Step 7: Check feasibility

Since \(m, n > 0\), each of \(m^2, n^2, mn\) is positive. Thus, \(m^2 + n^2 + mn > 0\), and cannot equal 0.

Final Conclusion:

No positive real numbers \(m, n\) can satisfy the given equation. \[ \boxed{\text{No solution exists.}} \]