SOLUTION: If {{{ log((x + y)/5)}}} = {{{ log(sqrt(x))+ log(sqrt(y)) }}} where x and y are greater than 0, show that {{{ x^2 + y^2 = 23xy }}}
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Question 614561: If = where x and y are greater than 0, show that
Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The sum of the logs is the log of the product, so:
\ +\ \log\left(\sqrt{x}\right)\ =\ \log\left(\sqrt{xy}\right))
Two log functions to the same base are equal if and only if their arguments are equal, hence:
Multiply both sides by 5. Square both sides. Add 
to both sides.
John
My calculator said it, I believe it, that settles it
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
If = where x and y are greater than 0, show that
**
log((x + y)/5)= log(sqrt(x))+ log(sqrt(y))
show that x^2 + y^2 = 23xy
..
log((x + y)/5)- log(sqrt(x)- log(sqrt(y)=0
log((x + y)/5)- [(log(sqrt(x)+ log(sqrt(y)]=0
place under single log:
log[((x + y)/5)/(√x*√y)]=0
convert to exponential form:
10^0=[((x + y)/5)/(√x*√y)=1
[((x + y)/5)=(√xy)
(x + y)=5√xy
square both sides
x^2+2xy+y^2=25xy
x^2+y^2=23xy
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