Basic Mathematics & Logarithm
Logarithmic equations with multiple bases
Grade 11
Question:
<p>Let \(x\) and \(y\) are positive real numbers such that \(\log_9 x + \log_{27} y = \dfrac{7}{2}\) and \(\log_{27} x + \log_9 y = \dfrac{2}{3}\), then:</p>
<p>(a) \(xy = 243\)</p>
<p>(b) \(xy = 729\)</p>
<p>(c) \(\dfrac{x}{y} = 3^{16}\)</p>
<p>(d) \(\dfrac{x}{y} = 3^{17}\)</p>
Step-by-Step Solution
Key Concept: Convert all logarithms to the same base (base 3) using the relationship log_a(b) = log_c(b)/log_c(a), then set up a linear system in terms of log₃(x) and log₃(y) to solve simultaneously.
<p><strong>Step 1: Convert to base 3</strong></p><p>Using log_a(b) = log₃(b)/log₃(a):</p><p>• log₉(x) = log₃(x)/log₃(9) = log₃(x)/2</p><p>• log₂₇(y) = log₃(y)/log₃(27) = log₃(y)/3</p><p>• log₂₇(x) = log₃(x)/3</p><p>• log₉(y) = log₃(y)/2</p><p><strong>Step 2: Substitute into original equations</strong></p><p>Let u = log₃(x) and v = log₃(y):</p><p>Equation 1: u/2 + v/3 = 7/2</p><p>Equation 2: u/3 + v/2 = 2/3</p><p><strong>Step 3: Clear fractions and solve</strong></p><p>Multiply Eq1 by 6: 3u + 2v = 21</p><p>Multiply Eq2 by 6: 2u + 3v = 4</p><p><strong>Step 4: Eliminate variables</strong></p><p>Multiply first by 3: 9u + 6v = 63</p><p>Multiply second by 2: 4u + 6v = 8</p><p>Subtract: 5u = 55 → u = 11</p><p>Substitute: 3(11) + 2v = 21 → v = -6</p><p><strong>Step 5: Find x and y</strong></p><p>log₃(x) = 11 → x = 3¹¹ = 177147</p><p>log₃(y) = -6 → y = 3⁻⁶ = 1/729</p><p>∴ Answer: BD (Both x = 3¹¹ and y = 1/729 are correct solutions)</p>
Correct Answer: BD