+0

# interesting question

0
72
3

Solve for x

$$x^{log_y{x}}=2$$

$$y^{log_x{y}}=16$$

x and y are greater than 0 of course

Mar 17, 2021

#1
+112953
+2

$$x^{log_y{x}}=2 \qquad \qquad \qquad y^{log_x{y}}=2^4\\ x^{\frac{log x}{log y}}=2 \qquad \qquad \qquad y^{\frac{log y}{log x}}=2^4\\ log\;{x^{\frac{log x}{log y}}}=log\;{2} \qquad \quad \log{y^{\frac{log y}{log x}}}=\log{2^4}\\ {\frac{log x}{log y}}*log\;x=log\;{2} \qquad \quad {\frac{log y}{log x}}*\log\;y=4\log\;2\\ \frac{({log x})^2}{log y}=log\;{2} \qquad \quad \qquad \frac{(log y)^2}{4log x}=\log\;2\\~\\ so\\ 4{{(log x})^3}=(log y)^3\\ \sqrt[3]{4}=\frac{logy}{logx}\\ log_xy = \sqrt[3]{4}$$

Oh I have to solve for x...

$$\frac{logx}{log y}=\frac{1}{\sqrt[3]{4}}\\ log_y x=4^{(-1/3)}\\ y^{log_y x}=y^{4^{(-1/3)}}\\ x=y^{4^{(-1/3)}}\\$$

Mar 17, 2021
#2
+32062
+3

Here's my solution to this:

Mar 17, 2021
#3
+1

my attempt:

Take logx of both sides

$$log_x{x^{log_y{x}}}=log_x2$$

(1) $$log_yx=log_x2$$

Take logy of both sides

$$log_y{y^{log_xy}}=log_y16$$

$$log_xy=log_y16$$

for (2) we isolate $$log_y2$$ on one side

$$log_xy=4*log_y2$$

(2) $$\frac{log_xy}{4}=log_y2$$

For (1) we invert the logs to substitute into 2

$$log_yx=log_x2$$ -> $$log_xy=log_2x$$

substituting:

$$\frac{log_2x}{4}=log_y2$$

Then we isolate y which is

$$y=2^{log_x16}$$

Substituing back into (1)
$$log_{2^{log_x16}}x=log_x2$$

$$\frac{log_2x}{log_x16}=log_x2$$

$$\frac{log_2x}{4*log_x2}=log_x2$$

$$log2_x=4*(log_x2)^2$$

We now make $$log_x2=a$$

$$\frac{1}{a}=4a^2$$

$$a=\frac{1}{4^{\frac{1}{3}}}$$

$$log_x2=\frac{1}{4^{\frac{1}{3}}}$$

solving for x we get x = 2^(cbrt(4))

Mar 17, 2021