if a^x = b^y = c^z and b^2 = ac then what is value of 1/x + 1/z?

Question

^ means to the power.

Answer 1

a^x = b^y = c^z ...1
a= b^(y/x) ...2
c= b^(y/z) ...3

ac = b^2 = b^(y/x + y/z) ... multiply 2and 3 and ac = b^2
so take log wrt b

2 = y/x + y/z or 1/x+ 1/z = 2/y

Answer 2

a^x = b^y = c^z
and b^2 = ac

→ a^x * c^z = b^y * b^y
→ a^x * c^z = b^2y
But b^2 = ac (given).
→ a^x * c^z = (ac)^y
→ a^x * c^z = a^y * c^y
→ x = y and z = y
So: x = y = z
--------------------------------
If x = y = z, then
→ a = b = c.
--------------------------------
1/x + 1/z = ?
But: x = z
→ 1/x + 1/z = 1/z + 1/z
1/x + 1/z = (z + z)/z²
1/x + 1/z = 2z/z²
1/x + 1/z = 2/z

But x = y = z

→ 1/x + 1/z = 2/z = 2/y = 2/x

Answer:
2/z (or) 2/y (or) 2/x

Answer 3

2/y

Answer 4

first consider the equality a^x=c^z
take logarithms on both the sides
log(a^x)=log(c^z)
a property of logarithms is log(p^q)=q*log(p)
so, log(a^x)=x*log(a) and log(c^z)=z*log(c)
so, xlog(a)=zlog(c)
=> z=x*(loga/logc)
to find 1/x+1/z
substitute the value of z in the above equation in terms of x
1/x+1/x(logc/loga)
=1/x((loga+logc)/loga)-----(1)
now, consider b^2=ac
apply logarithms on both the sides
so, 2logb=loga+logc
substitute in equation (1)
so, 1/x+1/z=1/x(2logb/loga)
=2/x(logb/loga)-----(2)
now consider a^x=b^y
apply logarithms again
xloga=ylogb
=>logb/loga = x/y
substitute this in equation (2)
so, 1/x+1/z=2/x(x/y)
=2/y
since b^2=ac, it implies that a,b and c are in a Geometric Progression (G.P)
since 1/x+1/z=2/y, it implies that x,y and z are in a Harmonic Progression (H.P)

Answer 5

please make your own homework