Advertisements
Advertisements
Question
If ax = by = cz and b2 = ac, show that `y=(2zx)/(z+x)`
Advertisements
Solution
Let ax = by = cz = k
So, `a=k^(1/x),` `b=k^(1/y),` c=k^(1/z)
Thus,
`b^2 = ac`
`rArr(k^(1/y))^2=(k^(1/x))(k^(1/z))`
`rArrk^(2/y)=k^(1/x+1/z)`
`rArr2/y=1/x+1/z`
`rArr2/y=(z+x)/(xz)`
`rArr2xx(zx)/(z+x)=y`
`rArry=(2zx)/(z+x)`
APPEARS IN
RELATED QUESTIONS
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
If `x=2^(1/3)+2^(2/3),` Show that x3 - 6x = 6
If `3^(x+1)=9^(x-2),` find the value of `2^(1+x)`
The seventh root of x divided by the eighth root of x is
The value of \[\left\{ \left( 23 + 2^2 \right)^{2/3} + (140 - 19 )^{1/2} \right\}^2 ,\] is
If g = `t^(2/3) + 4t^(-1/2)`, what is the value of g when t = 64?
If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x2 + y +y2 =
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
Simplify:
`7^(1/2) . 8^(1/2)`
