Advertisements
Advertisements
Question
If \[8^{x + 1}\] = 64 , what is the value of \[3^{2x + 1}\] ?
Options
1
3
9
27
Advertisements
Solution
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
RELATED QUESTIONS
If a = 3 and b = -2, find the values of :
aa + bb
Prove that:
`(x^a/x^b)^cxx(x^b/x^c)^axx(x^c/x^a)^b=1`
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
If 3x = 5y = (75)z, show that `z=(xy)/(2x+y)`
The value of x − yx-y when x = 2 and y = −2 is
If \[\frac{3^{5x} \times {81}^2 \times 6561}{3^{2x}} = 3^7\] then x =
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
Simplify:
`7^(1/2) . 8^(1/2)`
