Advertisements
Advertisements
Question
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Advertisements
Solution
`(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
RELATED QUESTIONS
Prove that:
`(a+b+c)/(a^-1b^-1+b^-1c^-1+c^-1a^-1)=abc`
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
Find the value of x in the following:
`5^(2x+3)=1`
The seventh root of x divided by the eighth root of x is
The value of 64-1/3 (641/3-642/3), is
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
If \[\sqrt{2} = 1 . 414,\] then the value of \[\sqrt{6} - \sqrt{3}\] upto three places of decimal is
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.
