If 2^x = 3^y = 6^z , then show that 1/x + 1/y = 1/z.
Answers
Step-by-step explanation:
Given:-
2^x = 3^y = 6^z
To find :-
Show that 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 the power 1/x both sides
=> (2^x)^1/x = k^(1/x)
=>2^(x/x) = k^(1/x)
Since (a^m)^n = a^mn
=> 2 = k^(1/x) -------------------(1)
On taking 3^y = k
On raising to the power 1/y both sides
=> (3^y)^1/y = k^(1/y)
=>3^(y/y) = k^(1/y)
Since (a^m)^n = a^mn
=> 3= k^(1/y) -------------------(2)
On taking 6^z = k
On raising to the power 1/z both sides
=> (6^z)^1/z = k^(1/z)
=>6^(z/z) = k^(1/z)
Since (a^m)^n = a^mn
=> 6 = k^(1/z) -------------------(3)
=> (2×3) = k^(1/z)
From (1)&(2)
=>k^(1/x)× k^(1/y) = k^(1/z)
=> k^(1/x)+(1/y) = k^(1/z)
Since a^m × a^n = a^(m+n)
Since the bases are equal then exponents must be equal
=> (1/x) +(1/y) = 1/z
Hence, Proved.
Used formulae:-
- (a^m)^n = a^mn
- a^m × a^n = a^(m+n)
- If the bases are equal then exponents must be equal
- If a^m = a^n => m = n