Advertisements
Advertisements
प्रश्न
Solve the following equation:
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
उत्तर
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
`rArr4^(2x)=(sqrt8)^2` and `(root3 16)^(-6/y)=(sqrt8)^2`
`rArr4^(2x)=(8^1/2xx2)` and `(16^(1/3xx-6/y))=(8^1/2xx2)`
`rArr4^(2x)=8` and `(16^(-2/y))=8`
`rArr2^(4x)=2^3` and `(2^(-8/y))=2^3`
`rArr4x=3` and `-8/y=3`
`rArrx=3/4` and `y=-8/3`
APPEARS IN
संबंधित प्रश्न
Find:-
`9^(3/2)`
Simplify the following:
`(6(8)^(n+1)+16(2)^(3n-2))/(10(2)^(3n+1)-7(8)^n)`
Solve the following equation for x:
`2^(x+1)=4^(x-3)`
Solve the following equation for x:
`2^(5x+3)=8^(x+3)`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
Find the value of x in the following:
`(3/5)^x(5/3)^(2x)=125/27`
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
Show that:
`((a+1/b)^mxx(a-1/b)^n)/((b+1/a)^mxx(b-1/a)^n)=(a/b)^(m+n)`
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
