Advertisements
Advertisements
प्रश्न
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Advertisements
उत्तर
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
LHS = `(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)`
`=(3^(a-b))^(a+b)(3^(b-c))^(b+c)(3^(c-a))^(c+a)`
`=(3^((a-b)(a+b)))(3^((b-c)(b+c)))(3^((c-a)(c+a)))`
`=(3^(a^2-b^2))(3^(b^2-c^2))(3^(c^2-a^2))`
`=3^(a^2-b^2+b^2-c^2+c^2-a^2)`
`=3^0`
= 1
= RHS
APPEARS IN
संबंधित प्रश्न
If `1176=2^a3^b7^c,` find a, b and c.
Simplify:
`root5((32)^-3)`
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
Prove that:
`(2^(1/2)xx3^(1/3)xx4^(1/4))/(10^(-1/5)xx5^(3/5))div(3^(4/3)xx5^(-7/5))/(4^(-3/5)xx6)=10`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Find the value of x in the following:
`5^(2x+3)=1`
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
Find:-
`32^(1/5)`
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.
