Advertisements
Advertisements
प्रश्न
If \[8^{x + 1}\] = 64 , what is the value of \[3^{2x + 1}\] ?
विकल्प
1
3
9
27
Advertisements
उत्तर
We have to find the value of `3^2x+1` provided `8^((x+1) = 64)`
So,
`2^(3(x+1)) = 64`
`2^(3x+3) = 2^6`
Equating the exponents we get
`3x + 3= 6 `
`3x= 6-3`
`3x=3`
`x= 3/3`
x - 1
By substitute in `3^(2x+1)`we get
`3^(2 xx 1 +1)`
` = 3^(2+1)`
`= 3^3`
`= 27`
The real value of `3^(2x+1)` is 27
APPEARS IN
संबंधित प्रश्न
If a = 3 and b = -2, find the values of :
ab + ba
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^-4/y^-10)^(5/4)`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Show that:
`[{x^(a(a-b))/x^(a(a+b))}div{x^(b(b-a))/x^(b(b+a))}]^(a+b)=1`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
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 \[\sqrt[5]{3125 a^{10} b^5 c^{10}}\] is equal to
The simplest rationalising factor of \[\sqrt{3} + \sqrt{5}\] is ______.
If x = \[\frac{2}{3 + \sqrt{7}}\],then (x−3)2 =
