How can be solved this problem with logarithm?

[Hmiemer]

$$\log_{xy^2}{xy}$$=? if $$\log_{x}{y}$$ $$-$$$$\log_{y}{x}$$ $$=$$ $$\frac{ 2 }{3 }$$answer: 3/5
i don‘t think that my way to solve this is at least somehow correct, though here it is, just in case​

 

[Hmiemer]

3/2 is correct! Not 2/3

 

[Dr.Peterson]

$$\log_{xy^2}{xy}$$=? if $$\log_{x}{y}$$ $$-$$$$\log_{y}{x}$$ $$=$$ $$\frac{ 2 }{3 }$$answer: 3/5
i don‘t think that my way to solve this is at least somehow correct, though here it is, just in case​

I think you're saying you want to find the value of $\log_{xy^2}{xy}$ if $$\log_{x}{y}-\log_{y}{x}=\frac{ 3 }{2}$$
You show some valid work, but in order to finish, you need to find the value of $\log_y x$, which I would do first.

Are you aware that $\log_y x = \frac{1}{\log_x y}$? That will be very useful when applied to the given equation.

 

[JeffM]

I think you're saying you want to find the value of $\log_{xy^2}{xy}$ if $$\log_{x}{y}-\log_{y}{x}=\frac{ 3 }{2}$$
You show some valid work, but in order to finish, you need to find the value of $\log_y x$, which I would do first.

Are you aware that $\log_y x = \frac{1}{\log_x y}$? That will be very useful when applied to the given equation.

To prove Dr. Peterson's theorem with a bit more specificity, notice that

$$\text {Given } \alpha = log_{\beta}(\gamma) \ne 0 \implies \beta > 0, \ \beta \ne 1,\ \gamma > 0, \text { and } \gamma \ne 1.$$
$$\alpha = log_{\beta}(\gamma) \implies \gamma = \beta^{\alpha} \text { by definition.}$$
$$\therefore log_{\gamma}(\gamma) = log_{\gamma}(\beta^{\alpha}) \implies 1 = \alpha * log_{\gamma}(\beta).$$
$$\therefore \text {If } log_{\gamma}(\beta) \ne 0, \text { then } \alpha = \dfrac{1}{log_{\gamma}(\beta)}.$$
$$log_{\gamma}(\beta) \ne 0 \text { because } \beta \ne 1 \text { by hypothesis.}$$
$$\therefore \alpha = \dfrac{1}{log_{\gamma}(\beta)} \implies log_{\beta}( \gamma) = \dfrac{1}{log_{\gamma}(\beta)} \text { if } log_{\beta}( \gamma) \ne 0.$$

 

[Dr.Peterson]

Are you aware that $\log_y x = \frac{1}{\log_x y}$?

This can be shown easily by just applying the change-of-base formula: $$\log_y x = \frac{log_x x}{\log_x y}= \frac{1}{\log_x y}$$ Of course, this assumes all the logs involved are defined, so that the formula applies.

 

[Steven G]

To find log_xy, I would let u = log_xy, then the equation becomes u - 1/u = 3/2. The solution to this can easily be seen or you can solve a quadratic equation.