Question

Let \(a=3\sqrt2\) and \(b=\frac{1}{5^{1/6}\sqrt5}\). If x, y ∈ R are such that
\(3x+2y=\log_a(18)^{\frac{5}{4}}\) and
\(2x-y=\log_b(\sqrt{1080})\), 
then 4x + 5y is equal to _______.

Answer 1

Correct Answer: 8

Given:

\[ a = 3\sqrt{2}, \quad b = \frac{1}{5^{1/6} \sqrt{6}}. \]  

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1. Evaluating \( \log_a \left( \frac{185}{4} \right) \)

Using the property:

\[ \log_a(b^n) = n \log_a(b), \]

we get:

\[ \log_a \left(\frac{185}{4} \right) = \frac{5}{4} \log_{3\sqrt{2}} (18). \]

Now, express \( \log_{3\sqrt{2}}(18) \):

\[ \log_{3\sqrt{2}}(18) = \frac{\log(18)}{\log(3\sqrt{2})}. \]

Since:

\[ \log(18) = 2\log(3) + \log(2), \quad \log(\sqrt{2}) = \frac{1}{2} \log(2), \]

we substitute:

\[ \log_{3\sqrt{2}}(18) = \frac{2 \log(3) + \log(2)}{\log(3) + \frac{1}{2} \log(2)}. \]

Numerically, this evaluates to:

\[ \log_a \left( \frac{185}{4} \right) = \frac{5}{2}. \]

Thus, we get the equation:

\[ 3x + 2y = \frac{5}{2}. \quad \text{(Equation 1)} \] 

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2. Evaluating \( \log_b (\sqrt{1080}) \)

Using the property:

\[ \log_{1/a}(b) = -\log_a(b), \]

we obtain:

\[ \log_b (\sqrt{1080}) = -\log_{5^{1/6} \sqrt{6}} (\sqrt{1080}). \]

Express \( \sqrt{1080} \) as:

\[ \sqrt{1080} = \sqrt{36 \times 30} = 6\sqrt{30}. \]

Now,

\[ \log_{5^{1/6} \sqrt{6}} (6\sqrt{30}) = \frac{\log(6\sqrt{30})}{\log(5^{1/6} \sqrt{6})}. \]

Simplifying:

\[ \log(6\sqrt{30}) = \log(6) + \frac{1}{2} \log(30). \]

Numerically, this evaluates to:

\[ \log_b (\sqrt{1080}) = -3. \]

Thus, we get the equation:

\[ 2x - y = -3. \quad \text{(Equation 2)} \] 

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3. Solving the equations:

From (1):

\[ 3x + 2y = \frac{5}{2}. \]

From (2):

\[ 2x - y = -3. \]

Multiply (2) by 2:

\[ 4x - 2y = -6. \]

Adding to (1):

\[ 3x + 2y + 4x - 2y = \frac{5}{2} - 6. \] \[ 7x = -\frac{7}{2} \Rightarrow x = -\frac{1}{2}. \]

Substituting \( x = -\frac{1}{2} \) into (2):

\[ 2 \times \left(-\frac{1}{2}\right) - y = -3. \] \[ -1 - y = -3 \Rightarrow y = 2. \] 

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4. Finding \( 4x + 5y \)

\[ 4x + 5y = 4 \times \left(-\frac{1}{2}\right) + 5(2). \] \[ = -2 + 10 = 8. \] 

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Final Answer:

\[ \mathbf{8} \]