Advertisements
Advertisements
Question
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
Advertisements
Solution
Let 2x = 3y = 6-z = k
`rArr2=k^(1/x),` `3=k^(1/y),` `6=k^(1/-z)`
Now,
`6 = 2xx3=k^(1/-z)`
`rArrk^(1/x)xxk^(1/y)=k^(1/-z)`
`rArrk^(1/x+1/y)=k^(1/-z)`
`rArr1/x+1/y=1/-z`
`rArr1/x+1/y+1/z=0`
APPEARS IN
RELATED QUESTIONS
Find:-
`9^(3/2)`
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
Find the value of x in the following:
`5^(x-2)xx3^(2x-3)=135`
Solve the following equation:
`8^(x+1)=16^(y+2)` and, `(1/2)^(3+x)=(1/4)^(3y)`
If a and b are different positive primes such that
`(a+b)^-1(a^-1+b^-1)=a^xb^y,` find x + y + 2.
If 1176 = `2^axx3^bxx7^c,` find the values of a, b and c. Hence, compute the value of `2^axx3^bxx7^-c` as a fraction.
If 3x-1 = 9 and 4y+2 = 64, what is the value of \[\frac{x}{y}\] ?
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
If \[\frac{5 - \sqrt{3}}{2 + \sqrt{3}} = x + y\sqrt{3}\] , then
Simplify:
`(3/5)^4 (8/5)^-12 (32/5)^6`
