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(1/27) ^ 3x+2 (simplify)
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\[(\frac{1}{27})^{3x+2}=\frac{1^{3x+2}}{27^{3x+2}}\]\[=\frac{1^{3x}+1^2}{27^{3x+2}}\]\[=\frac{1+1}{27^{3x+2}}\]\[=\frac{2}{27^{3x+2}}\]
@Zed, the numerator is just 1, not 2, since you wrote \[1^{3x+2}=1^{3x}+1^2\]
why did u remove the exponents of 1?
Yes it is, thanks imperialist. My mistake :)
@shinigami1m: 1 to any power is just 1
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It should be \[\frac{1}{27^{3x+2}}\]
it can be zero
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
oh
(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)
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