Question 1 Report
If \(2^{2x - 3y} = 32\) and \(\log_{y} x = 2\), find the values of x and y.
Given \(2^{2x-3y}=32\) and \(\log_y x=2.\)
First equation. Since \(32=2^5,\) equate indices:
\[2x-3y=5.\quad(1)\]
Second equation. \(\log_y x=2\) means \(x=y^2.\quad(2)\)
Substitute (2) into (1):
\[2y^2-3y=5\Rightarrow 2y^2-3y-5=0.\]
Factorise: \((2y-5)(y+1)=0,\) so \(y=\dfrac{5}{2}\) or \(y=-1.\)
A logarithm base must be positive and not equal to 1, so \(y=-1\) is rejected. Hence
\[y=\frac{5}{2},\qquad x=y^2=\frac{25}{4}.\]
Check: \(2x-3y=2\!\left(\tfrac{25}{4}\right)-3\!\left(\tfrac{5}{2}\right)=\tfrac{25}{2}-\tfrac{15}{2}=5.\) Correct.
Answer Details
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