SOLUTION: Simultaneous equations log base 4 xy = 10 2 log base 8 x = 3 log base 8 y

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

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