If \(2^{2x - 3y} = 32\) and \(\log_{y} x =…

Question 1 Report

If \(2^{2x - 3y} = 32\) and \(\log_{y} x = 2\), find the values of x and y.

Answer

Answer Details

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.

If \(2^{2x - 3y} = 32\) and \(\log_{y} x = 2\), find the values of x and y.