Answer:



The given equation is a*x = b*y = c*z and xyz are in geometric progression (GP).



We know that if a, b, c are in GP, then loga, logb, logc are in AP (arithmetic progression).

Let's first prove that a, b, c are in GP.

We can write the equation as:

a = (b*y)/(x) = (c*z)/(x*y)

b/a = x/y

c/b = y/z

Therefore, b/a = c/b = (y/x)*(z/y) = z/x

Hence, a, b, c are in GP.

Now, we can say that loga, logb, logc are in AP.

Proof:

Let loga = A, logb = B, logc = C

We need to prove that B - A = C - B

From the given equation, we can write:

x = b*y/a

y = c*z/b

Substituting these in xyz = b*y*c*z/a*b, we get:

xyz = c*z*x

Taking logarithm on both sides with base a, we get:

loga(xyz) = loga(c*z*x)

loga(x) + loga(y) + loga(z) = loga(c) + loga(z) + loga(x)

A+B+C = loga(bc) + loga(xz)

A+B+C = loga(bcxz)

Similarly, taking logarithm with base b and c, we get:

A+B+C = logb(acxz) = logc(abxz)

Therefore, loga, logb, logc are in AP with common difference (B - A) = (C - B).



Hence, we can conclude that if a*x = b*y = c*z and xyz are in GP, then loga, logb, logc are in AP.