logx+log(xy8)(logx)2+(logy)2=2,logy+log(x8y)(logx)2+(logy)2=0 \large \log x + \dfrac{ \log(xy^8)}{(\log x)^2 + (\log y)^2 } = 2, \qquad \log y + \dfrac{ \log \left( \dfrac{x^8}y \right)}{(\log x)^2 + (\log y)^2 } = 0 logx+(logx)2+(logy)2log(xy8)=2,logy+(logx)2+(logy)2log(yx8)=0
Let xxx and yyy satisfy the system of equations above. Find the product xyxyxy.
Note by Naitik Sanghavi 5 years, 1 month ago
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Is ithe dot multiplication sign or subtraction?
If multiplication , then take ligx=a and logy=b and then proceed.It would be easier now.
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Its Logx+log(xy^8)/(logx)²+(logy)²=2
Logy+log(x^8y)/(logx)²+(logy)²=0
Where is the answer
@Pi Han Goh ,Thanks!
Can you help me solve this!?
@Calvin Lin @Nihar Mahajan @Swapnil Das @Svatejas Shivakumar @Sravanth Chebrolu @Brian Charlesworth @Rajdeep Dhingra @Mehul Arora can someone help me!
Please limit your @mentions to at most 5 people.
Sorry for that,but still no one replied !!
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
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or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Is ithe dot multiplication sign or subtraction?
If multiplication , then take ligx=a and logy=b and then proceed.It would be easier now.
Log in to reply
Its Logx+log(xy^8)/(logx)²+(logy)²=2
Logy+log(x^8y)/(logx)²+(logy)²=0
Log in to reply
Where is the answer
@Pi Han Goh ,Thanks!
Can you help me solve this!?
@Calvin Lin @Nihar Mahajan @Swapnil Das @Svatejas Shivakumar @Sravanth Chebrolu @Brian Charlesworth @Rajdeep Dhingra @Mehul Arora can someone help me!
Log in to reply
Please limit your @mentions to at most 5 people.
Log in to reply
Sorry for that,but still no one replied !!