Question

(1/27) ^ 3x+2 (simplify)

Answer 1

\[(\frac{1}{27})^{3x+2}=\frac{1^{3x+2}}{27^{3x+2}}\]\[=\frac{1^{3x}+1^2}{27^{3x+2}}\]\[=\frac{1+1}{27^{3x+2}}\]\[=\frac{2}{27^{3x+2}}\]

Answer 2

@Zed, the numerator is just 1, not 2, since you wrote \[1^{3x+2}=1^{3x}+1^2\]

Answer 3

why did u remove the exponents of 1?

Answer 4

Yes it is, thanks imperialist. My mistake :)

Answer 5

@shinigami1m: 1 to any power is just 1

Answer 6

It should be \[\frac{1}{27^{3x+2}}\]

Answer 7

it can be zero

Answer 8

No it cannot. Even 1 to the zeroth power is still 1. 1 raised to any complex number (so including the reals) is always 1

Answer 9

oh

Answer 10

(1/27)^2x+2
(1^3x*1^2)/27^(3x+2);
{1^(3x+2)+1]/[3^(9x+6);
(1+1)/[3^(9x+6);
final ans:
2/[3^(9x+6)