Question

show that (9^(n+1))/(3^(2n-1)) has a value of 27

Answer 1

9 = 3^2

so 9^(n+1) = (3^2)^(n+1) = 3^(2n+2)

Keep in mind that (n^a)^b = n^(ab)

Now take 3^ (n+2) and divide it by 3^(2n-1)..using the law of exponents...a^m/a^n = a^(m-n)

Answer 2

\[x^{a+b} = x^ax^b\]\[(x^a)^{bc} = x^{abc} = (x^{ab})^c = etc.\]

\[\frac{ 9^{n+1} }{ 3^{2n - 1} } = \frac{ 9^n * 9^1 }{ 9^n*3^{-1} } = \frac{ 9 }{ \frac{ 1 }{ 3} } = 27\]