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 :
aa + bb
Simplify:
`root5((32)^-3)`
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
If `27^x=9/3^x,` find x.
Find the value of x in the following:
`2^(x-7)xx5^(x-4)=1250`
Solve the following equation:
`3^(x+1)=27xx3^4`
State the power law of exponents.
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
If o <y <x, which statement must be true?
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
