Question 944178: Simultaneous equations
log base 4 xy = 10
2 log base 8 x = 3 log base 8 y
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i'll show log base 4 as log4
i'll show log base 8 as log8
your equations are:
log4(xy) = 10
2log8(x) = 3log8(y)
since, in general, a*log(b) = log(b^a), your second equation can be simplified to:
log8(x^2) = log8(y^3)
this can only be true if x^2 = y^3
solve for y to get y = x^(2/3)
go back to your first equation and replace y with x^(2/3) to get:
log4(xy) = 10 becomes log4(x * x^(2/3)) = 10
since, in general, x^a * x^b equals x^(a+b), x * x^(3/3) becomes x^(5/3).
your equation becomes:
log4(x^(5/3)) = 10
this is true if and only if 4^10 = x^(5/3)
solve for x to get:
x = (4^10)^(3/5) which simplifies to:
x = 4^(30/5) which further simplifies to 4^6 which is equal to 4096.
you have x = 4096.
since y = x^(2/3), then y = 4096^(2/3) which makes y = 256.
you have x = 4096 and y = 256.
go back to your first equation of log4(xy) = 10 and replace x with 4096 and y with 256 to get:
log4(xy) = 10 becomes log4(4096*256) = 10.
this is true if and only if 4^10 = 4096*256.
simplify to get 1048576 = 1048576 so you're good.
you can also confirm that 2log8(x) = 3log(y) by replacing x with 4096 and y with 256 to get:
2log8(4096) = 3log(256)
you can use the log conversion to the base 10 formula to see if this is true.
2log8(4096) = 3log(256) becomes 2log10(4096)/log10(8) = 3log10(256)/log10(8) which you can solve using your calculator LOG function to get 8 = 8.
this confirms the values for x and y are good.
x = 4096
y = 256
|
|
|
