How do i solve for Z?

Question

log base 2x (Z) = 3
log base 5y (Z) = 6
log base xy (Z) = 2/3

how do i solve for Z?

Answer 1

Given:

log base 2x (Z) = 3
log base 5y (Z) = 6
log base xy (Z) = 2/3

Exponentiate

z = (2x)^3
z = (5y)^6
z = (xy)^(2/3)

Setting the first and second equations equal.

(2x)^3 = (5y)^6
2x = (5y)^2 = (5^2)y^2
x = ((5^2)/2)y^2

Setting the first and third equations equal.

(2x)^3 = (xy)^(2/3)
(2x)^9 = (xy)^2
(2^9)(x^9) = (x^2)(y^2)
x^7 = (y^2)/(2^9)
x = [y^(2/7)/2^(9/7)]

Setting the two values of x equal.

((5^2)/2)y^2 = [y^(2/7)/2^(9/7)]
y^(12/7) = 1/{(5^2)(2^(2/7))}
y = 1/{(5^2)(2^(2/7))}^(7/12)
y = 1/{[5^(7/6)][2^(1/6)]}
y = (1/5){1/{[5^(1/6)][2^(1/6)]}
y = (1/5)(1/10)^(1/6)

Plugging into the second equation for z.

z = (5y)^6 = (5^6)(y^6) =
z = (5^6){(1/5)(1/10)^(1/6)}^6 = 1/10

Answer 2

Perform an inverse log operation on each side (take each side and make it a power of the base that the log is based in, so log base x of y = a would become x^(log base x of y) = x^a which would become y = x^a ). After that, you have three equations and three unknowns and should be able to solve each of them.