Question

please help me to solve the problem...With full process....​

Answer 1

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

⭐Actually Welcome to the Concept of the Logarithms and it's properties

⭐Basically here we are going to use the property that's

⭐log a base b = log a/ log b

and also

⭐Log m raised to power n = n log m

Answer 2

Answer:

Let:

log1/b(a) = x => 1/b^x = a … (1)

log1/c(b) = y => 1/c^y = b … (2)

log1/a(c) = z => 1/a^z = c … (3)

Then by (1), (2) and (3), it follows that:

a =1/b^x

=> a = (1/c^y)^x

=> a = 1/c^(xy)

=> a = (1/a^z)^(xy)

=> a^1 = 1/a^(xyz)

=> a^1 = a^(-xyz)

=> xyz = -1 …....................(4)

Therefore, by (4) we take:

[log1/b(a)][log1/c(b)[log1/a(c)] = (-1) (Ans.)

( This one is the best and easiest method I think ,

I didn't understand other solution that was given here)

$\color{red}{Mark\: as\: Brainlist}$