Alg II / Pre-Calc w/ Logs: Given log_x(y)+log_y(x)=3, find log_x(y^2)+log_y(x^2)

kikithecat

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Hi there.

The question I have to complete is this : "Let x and y be two positive real numbers such that logxy+logyx = 3. Find the value of logxy2+logyx2." Now, I have simplified the equation down to 2logx(y)+2logy(x), but I don't know how to find the value after that. Would the 3 from the original equation have to be squared as well? Any guidance from there would be helpful.

Thanks!
 
Hi there.

The question I have to complete is this : "Let x and y be two positive real numbers such that logxy+logyx = 3. Find the value of logxy2+logyx2." Now, I have simplified the equation down to 2logx(y)+2logy(x), but I don't know how to find the value after that. Would the 3 from the original equation have to be squared as well? Any guidance from there would be helpful.

Thanks!

2logx(y)+2logy(x) ......... factor out 2

= 2 * [logx(y)+logy(x)] ...... continue....
 
Hi there.

The question I have to complete is this : "Let x and y be two positive real numbers such that logxy+logyx = 3. Find the value of logxy2+logyx2." Now, I have simplified the equation down to 2logx(y)+2logy(x), but I don't know how to find the value after that. Would the 3 from the original equation have to be squared as well? Any guidance from there would be helpful.

Thanks!

Don't square; just factor out the 2!
 
Hi there.

The question I have to complete is this : "Let x and y be two positive real numbers such that logxy+logyx = 3. Find the value of logxy2+logyx2." Now, I have simplified the equation down to 2logx(y)+2logy(x), but I don't know how to find the value after that. Would the 3 from the original equation have to be squared as well? Any guidance from there would be helpful.

Thanks!
Try it this way.

\(\displaystyle \text {GIVEN: } x,\ y \in \mathbb R^+ \text { and } log_x(y) + log_x(y) = 3.\)

\(\displaystyle \text {Let } u = log_x(y) \text { and } v = log_y(x) \implies\)

\(\displaystyle u + v = 3 \text { and } log_x(y^2) = 2u \text { and } log_y(x^2) = 2v \implies \)

\(\displaystyle log_x(y^2) + log_y(x^2) = 2u + 2v = 2(u + v) = WHAT?\)
 
Try it this way.

\(\displaystyle \text {GIVEN: } x,\ y \in \mathbb R^+ \text { and } log_x(y) + log_x(y) = 3.\)

\(\displaystyle \text {Let } u = log_x(y) \text { and } v = log_y(x) \implies\)

\(\displaystyle u + v = 3 \text { and } log_x(y^2) = 2u \text { and } log_y(x^2) = 2v \implies \)

\(\displaystyle log_x(y^2) + log_y(x^2) = 2u + 2v = 2(u + v) = WHAT?\)

Oh, I see now! Using a simpler version of the problem was helpful. I was over-analyzing it, believing that I needed to find the values of x and y when it was really as simple as that. Thank you!
 
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