2^x=3^y=6^z then find the value of 1/x+1/y+1/z
Answers
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Step-by-step explanation:
Given :-
2^x=3^y=6^z
Correction:-
2^x = 3^y = 6^-z
To find :-
Find the value of 1/x+1/y+1/z ?
Solution :-
Given that
2^x = 3^y = 6^-z
Let 2^x = 3^y = 6^-z = k
On taking 2^x = k
On raising to 1/x both sides then
=> (2^x)^(1/x) = k^(1/x)
=> 2^(x/x) = k^(1/x)
=> 2^1 = k^(1/x)
=> 2 = k^(1/x) ---------------(1)
and
On taking 3^y = k
On raising to 1/y both sides then
=> (3^y)^(1/y) = k^(1/y)
=> 3^(y/y) = k^(1/y)
=> 3^1 = k^(1/y)
=> 3 = k^(1/y) ---------------(2)
On taking 6^-z = k
On raising to 1/z both sides then
=> (6^-z)^(1/z) = k^(1/z)
=> 6^(-z/z) = k^(1/z)
=> 6^-1 = k^(1/z)
=> 6^-1 = k^(1/z)
=>(6^-1)^-1 = [k^(1/z)]^-1
=> 6 = k^(-1/z)
=> [(2×3)]= k^(-1/z)
=> [{k^(1/x)}×{k^(1/y)}] = k^(-1/z)
=> [k^{(1/x)+(1/y)}]= k^(-1/z)
Since , If bases are equal then exponents must be equal.
=> [(1/x)+(1/y)]= -1/z
=> (1/x)+(1/y)+(1/z) = 0
Answer:-
The value of (1/x)+(1/y)+(1/z) for the given problem is 0
Used formulae:-
→ If bases are equal then exponents must be equal.
→ (a^m)^n = a^(mn)
→ a^m × a^n = a^(m+n)
→ a^-n = 1/a^n