If 2^x = 3^y =6^-z ,then show that 1/x +1/y +1/z=0
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Answers
Step-by-step explanation:
If 2^x = 3^y =6^-z ,then show that 1/x +1/y +1/z=0
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Step-by-step explanation:
Given :-
2^x = 3^y =6^-z
To find :-
Show that 1/x +1/y +1/z=0
Solution :-
Method -1:-
Given that :
2^x = 3^y =6^-z
Let 2^x = 3^y =6^-z = k (say)
On taking 2^x = k
On raising power to 1/x both sides then
=> (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 power to 1/y both sides then
=> (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 power to 1/z both sides then
=> (6^-z)^(1/z) = k^(1/z)
=> 6^(-z/z) = k^(1/z)
Since (a^m)^n = a^mn
=> 6^-1 = k^(1/z)
=> 1/6 = k^(1/z)
=> k^(1/z) = 1/6
=> 6×k^(1/z) = 1
=> (2×3) ×k^(1/z) = 1
=> k^(1/x) × k^(1/y) × k^(1/z) = 1
=> k^(1/x + 1/y + 1/z) = 1
=> k^(1/x + 1/y + 1/z) = k⁰
Since a⁰ = 1
If the bases are equal then exponents must be equal
=> 1/x + 1/y + 1/z = 0
Hence, Proved.
Method -2:-
Given that :
2^x = 3^y =6^-z
Let 2^x = 3^y =6^-z = k (say)
On taking 2^x = k
On taking logarithms both sides then
=> log 2^x = log k
=> x log 2 = log k
Since log a^m = m log a
=> x = log k / log 2
=> 1/x = log 2 / log k -------(1)
On taking 3^y = k
On taking logarithms both sides then
=> log 3^y = log k
=> y log 3= log k
Since log a^m = m log a
=> y = log k / log 3
=> 1/y = log 3 / log k -------(2)
On taking 6^-z = k
On taking logarithms both sides then
=> log 6^-z= log k
=>-z log 6 = log k
Since log a^m = m log a
=> -z = log k / log 6
=> z =-( log k / log 6)
=> 1/z= -(log 6 / log k) -------(3)
On adding (1),(2)&(3)
1/x + 1/y + 1/z
=> (log 2/log k) +(log 3/log k) -(log 6/log k)
=> (log 2+ log 3 - log 6)/log k
=> [log (2×3)- log 6]/log k
Since log a + log b = log ab
=> (log 6-log 6)/log k
=> 0/log k
=> 0
Hence, Proved.
Answer :-
If 2^x = 3^y =6^-z then 1/x +1/y +1/z=0
Used formulae:-
- (a^m)^n = a^mn
- a⁰ = 1
- log a^m = m log a
- log a + log b = log ab