Question

If a^1/x = b^1/y = c^1/z and abc = 1 prove that x+y+z = 0​

Answer 1

Answer:

1-1=0

Step-by-step explanation:

step-by-step explanation

Answer 2

Question :- if a^(1/x) = b^(1/y) = c^(1/z) , and abc = 1 , prove that, x + y + z = 0 .

Solution :-

Let us assume that,

→ a^(1/x) = b^(1/y) = c^(1/z) = k , (where k is a constant number.)

then,

→ a^(1/x) = k

→ a = k^x

similarly,

→ b^(1/y) = k

→ b = k^y

and,

→ c^(1/z) = k

→ c = k^z .

given that,

→ a * b * c = 1 .

putting above values we get,

→ k^x * k^y * k^z = 1

using a^l * a^m * a^n = a^(l + m + n) , we get,

→ k^(x + y + z) = 1

now, we know that, if power of a number is 0 , it is equal to 1,

therefore,

→ k^(x + y + z) = k^0

now, we know that, when base is same, powers are equal . { a^m = a^n => m = n }

hence,

→ (x + y + z) = 0 (Proved.)

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