Question 944178
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