Question

Let a, b, c be three distinct positive real numbers such that (2a)^log_ea = (bc)^log_eb and b^log_e2 = a^log_ec. Then 6a + 5bc is equal to _____.

Answer 1

Correct Answer: 8

Given the equation:

\[ (2a) \ln a = (bc) \ln b, \quad 2a > 0, \quad bc > 0 \]

Step 1: Taking the Natural Logarithm 

Rewriting the equation:

\[ \ln a (\ln 2 + \ln a) = \ln b (\ln b + \ln c) \]

Let:

• \( \ln 2 = \alpha \)
• \( \ln a = x \)
• \( \ln b = y \)
• \( \ln c = z \)

The equations become:

• \( \alpha y = xz \)
• \( x(\alpha + x) = y(y + z) \)

Step 2: Rearranging the Equations

From \( \alpha y = xz \), we get:

\[ \alpha = \frac{xz}{y} \]

Substituting into the second equation:

\[ x^2 (z + y) = y^2 (y + z) \]

Factorizing gives:

\[ (y + z)(x^2 - y^2) = 0 \]

This implies either:

• \( y + z = 0 \) or
• \( x^2 = y^2 \), which gives \( x = -y \).

Step 3: Case 1 (\( bc = 1 \))

If \( bc = 1 \), then:

\[ (2a) \ln a = 1 \]

Solving for \( a \):

\[ a = 1 \quad \text{or} \quad a = \frac{1}{2} \]

For \( a = 1 \), the solution is:

\[ (a, b, c) = \left( \frac{1}{2}, \lambda, \frac{1}{\lambda} \right), \quad \lambda = 1, 2, \frac{1}{2} \]

Step 4: Case 2 (\( ab = 1 \))

If \( ab = 1 \), then the solutions are:

\[ (a, b, c) = \left( \lambda, \frac{1}{\lambda}, \frac{1}{2} \right), \quad \lambda = 1, 2, \frac{1}{2} \]

Step 5: Summing the Values

For the given conditions, summing \( 6a + 5bc \):

\[ 6a + 5bc = 3 + 5 = 8 \]

Final Answer:

\[ \boxed{8} \]