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 }}}

Algebra ->  Exponential-and-logarithmic-functions -> 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 }}}      Log On


   

Question 614561: If +log%28%28x+%2B+y%29%2F5%29 = +log%28sqrt%28x%29%29%2B+log%28sqrt%28y%29%29+ where x and y are greater than 0, show that +x%5E2+%2B+y%5E2+=+23xy+
Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The sum of the logs is the log of the product, so:



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

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Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
If +log%28%28x+%2B+y%29%2F5%29 = +log%28sqrt%28x%29%29%2B+log%28sqrt%28y%29%29+ where x and y are greater than 0, show that +x%5E2+%2B+y%5E2+=+23xy+
**
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