SOLUTION: Please help me solve for x and y: (log(3,x))(log(x,2x))(log(2x,y)) = log(x,y^2) Thanks

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please help me solve for x and y: (log(3,x))(log(x,2x))(log(2x,y)) = log(x,y^2) Thanks      Log On


   



Question 549741: Please help me solve for x and y:
(log(3,x))(log(x,2x))(log(2x,y)) = log(x,y^2)
Thanks

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i think you need to convert everything to the log with the same base.
i used base of 10.
when i did that, the y disappeared and i was able to solve for x to get x = 9.
that's the solution to this equation as far as i can see.
x = 9
y can be anything.
your starting equation is:
log(3,x)*log(x,2x)*log(2x,y) = log(x,y^2)
convert everything to the base of 10 using the conversion formula of:
log(b,x) = log(x)/log(b)
b is the base.
x is the thing you want to take the log of.
example:
log(3,x) is the log of x to the base of 3.
to convert log(3,x) to a log with the base of 10, you apply the conversation formula to get:
log(3,x) = log(x)/log(3).
applying the formula to the rest of the equation, you get:
log(x)/log(3) * log(2x)/log(x) * log(y)/log(2x) = log(y^2)/log(x)
log(x) and log(2x) in the expression on the left side of the equation cancel out to get:
log(y)/log(3) = log(y^2)/log(x)
since log(y^2) = 2log(y), this equation becomes:
log(y)/log(3) = 2log(y)/log(x)
divide both sides of this equation by log(y) to get:
1/log(3) = 2/log(x)
multiply both sides of this equation by log(3) and then multiply both sides of this equation by log(x) to get:
log(x) = 2log(3)
since we're dealing in the base of 10, we can use our calculator to solve for log(x) to get:
log(x) = .9542425094
use the anti-log formula in your calculator to solve for x to get:
x = 9
you can confirm the answer is correct by substituting 9 for x and substituting any other value for y and you will see that the original equation holds true.
i did and i got a true equation so i'm pretty sure i'm correct.