Answer to Question #222481 in Algebra for Chrissy

Question

Solve the simultaneous equations

2^{x^2} = 4^y and log_xy = 3/2 

Answer 1

Explanations & Calculations

$\qquad\qquad \begin{aligned} \small 2^{x^2}&=\small 4^y\\ \small 2^{x^2}&=\small (2^2)^y= 2^{2y}\\\\ \text{By equaling the indeces we get}\\\\ \small x^2&=\small 2y\cdots\cdots\cdots\cdots(1)\\\\ \small log_x\,y&=\small \frac{3}{2}\\ \small y&=\small x^{\frac{3}{2}}\cdots\cdots\cdots\cdots(2)\\\\ \small\text{By (1)=(2),}\\\\ \small x^2 &=\small 2x^{\frac{3}{2}}\\ \small x^2-2x^{\frac{3}{2}}&=\small 0\\ x^{\frac{3}{2}}(x^{\frac{1}{2}}-2)&=\small 0\\ \small x^{\frac{3}{2}} = 0\qquad &or\qquad x^{\frac{1}{2}}-2=0\\ \small x= 0\qquad &or\qquad x =2^2=4 \\\\ \small \text{accordingly the y values by (2),}\\\\ \small y = 0\qquad & or \qquad y= 8\\\\ \small \therefore (x,y) = (0,0)\qquad &or\qquad (4,8) \end{aligned}$

• All x & y values satisfy the above (1) & (2) equations hence form the results.