Advertisements
Advertisements
प्रश्न
Prove the following:
`root("ab")(x^"a"/x^"b")·root("bc")(x^"b"/x^"c")·root("ca")(x^"c"/x^"a")` = 1
Advertisements
उत्तर
L.H.S.
= `root("ab")(x^"a"/x^"b")·root("bc")(x^"b"/x^"c")·root("ca")(x^"c"/x^"a")`
= `(x^"a"/x^"b")^(1/"ab")·(x^"b"/x^"c")^(1/"bc")·(x^"c"/x^"a")^(1/"ca")`
= `x^(1/"b")/(x1/"a")·x^(1/"c")/(x1/"b")·x^(1/"a")/(x1/"c")` .....(Using (am)n = amn)
= `x^(1/"b"-1/"a")·x^(1/"c"-1/"b")·x^(1/"a"-1/"c")` ....(Using am ÷ an = am-n)
= `x^(("a"-"b")/"ab").x^(("b"-"c")/"bc").x^(("c"-"a")/"ac")`
= `x^(("a-b")/("ab")+("b-c")/("bc")+("c-a")/("ac")` ....(Using am x an = am+n)
= `x^(("ac"-"bc"+"ab"-"ac"+"bc"-"ab")/"abc"`
= `x^(0/"abc")`
= x0
= 1 ......(Using a0 = 1)
=R.H.S.
Hence proved.
APPEARS IN
संबंधित प्रश्न
Solve : 8 x 22x + 4 x 2x + 1 = 1 + 2x
Solve for x:
`2^(3x + 3) = 2^(3x + 1) + 48`
Solve : 3x-1× 52y-3 = 225.
If 2x = 4y = 8z and `1/(2x) + 1/(4y) + 1/(8z) = 4` , find the value of x.
Evaluate the following:
`(2^3 xx 3^5 xx 24^2)/(12^2 xx 18^3 xx 27)`
Solve for x:
`9 xx 3^x = (27)^(2x - 5)`
Solve for x:
2x + 3 + 2x + 1 = 320
Solve for x:
22x+3 - 9 x 2x + 1 = 0
Solve for x:
22x + 2x +2 - 4 x 23 = 0
Prove the following:
`(x^("a"+"b")/x^"c")^("a"-"b") · (x^("c"+"a")/(x^"b"))^("c"-"a") · ((x^("b"+"c"))/(x"a"))^("b"-"c")` = 1
