How can be solved this problem with logarithm?

[Hmiemer]

[math]\log_{xy^2}{xy}[/math]=? if [math]\log_{x}{y}[/math] [math]-[/math][math]\log_{y}{x}[/math] [math]=[/math] [math]\frac{ 2 }{3 }[/math]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]

[math]\log_{xy^2}{xy}[/math]=? if [math]\log_{x}{y}[/math] [math]-[/math][math]\log_{y}{x}[/math] [math]=[/math] [math]\frac{ 2 }{3 }[/math]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 [imath]\log_{xy^2}{xy}[/imath] if [math]\log_{x}{y}-\log_{y}{x}=\frac{ 3 }{2}[/math]
You show some valid work, but in order to finish, you need to find the value of [imath]\log_y x[/imath], which I would do first.

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

 

[JeffM]

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

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

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

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

 

[Dr.Peterson]

Are you aware that [imath]\log_y x = \frac{1}{\log_x y}[/imath]?

This can be shown easily by just applying the change-of-base formula: [math]\log_y x = \frac{log_x x}{\log_x y}= \frac{1}{\log_x y}[/math] 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.