SOLUTION: I am unable to solve this problem. Kindly help.
Log10(x+y)/3 = 1/2(log10x + log10y) Now, prove that x/y + y/x = 7
Algebra.Com
Question 5851: I am unable to solve this problem. Kindly help.
Log10(x+y)/3 = 1/2(log10x + log10y) Now, prove that x/y + y/x = 7
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
before anyone can help, you need to explain a couple of things:
1. is the 10 the base of the log?
2. the "/3" is this contained within the log ie log((x+y)/3) or outside, as in (log(x+y))/3. Your notation does not make it clear.
cheers
Jon.
RELATED QUESTIONS
I need help with this problem. Directions is solve for y in terms of x.
log10y = log10... (answered by Alan3354)
Solve for x:
log10 7 = log10x - log10... (answered by Fombitz)
I have two questions that I am confused on Please help!
1. Solve for x: log10 9 =... (answered by ankor@dixie-net.com)
log10 x * log10... (answered by Alan3354)
Hello,
I need help with a few problems i cant seem to work out for homework.... (answered by AnlytcPhil)
given that log10^7=x and log10^2=y, evaluate... (answered by Theo,Edwin McCravy)
If x=log15 2, y= log10 3, z=log6 5; prove that xy + yz + zx + 2xyz =... (answered by robertb)
SOLUTION: If x=log15 2, y= log10 3, z=log6 5; prove that xy + yz + zx + 2xyz =... (answered by ikleyn)
Please help me solve this problem:
y=1/2x-3
y=1/3x+2
I tried this
y=1/3x+2... (answered by stanbon)