Question

If a^x = b^y = c^z and b^2 = ac, then find 1/x + 1/z.

Answer 1

a^x = c^z
x log(a) = z log(c)
x log(a) + z log(a) = z log(c) + z log(a)
(x+z) log(a) = z ( log (a) + log(c) )
(x+z) log(a) = z log (ac)
(x+z) log(a) = z log (b^2) as b^2=ac
(x+z) log(a) = 2z log (b)

as a^x = b^y,
x log(a) = y log(b)
log (b) = (x/y) * log(a)

hence,
(x+z) log(a) = 2z*(x/y) * log(a)
(x+z) = 2xz/y
(x+z)/xz = 2/y
1/x + 1/z = 2/y

Answer 2

Does this have to involve logarithms? Because I'm just in ninth grade.

Answer 3

Another method is

Let ( a^x ) = ( b^y ) = ( c^z ) = k.

hence, a = k^(1/x), b = k^(1/y), c = k^(1/z).

as b² = ac.

[ k^(1/y) ]² = [ k^(1/x) ]·[ k^(1/z) ]
[ k^(2/y) = k^[ (1/x) + (1/z) ]

2/y = (1/x) + (1/z)
2/y = ( z+x) / (xz)
( z+x) / (xz) = 2/y
1/x + 1/z = 2/y

what about this ?

Answer 4

Thanks!!