Bunuel wrote:
If \(\frac{4}{3^9} = a - \frac{2}{3^7} - \frac{2}{3^8} - \frac{2}{3^9}\), then what is the value of a?
A. 32/3^33
B. 10/3^8
C. 10/3^6
D. 10/3^2
E. 3
If you don't see
niks18 's shortcut, and if you keep your wits about you, this method takes much less than the average time:
\(\frac{4}{3^9} = a - \frac{2}{3^7} - \frac{2}{3^8} - \frac{2}{3^9}\)
**Multiply all terms by \(3^9\) =
\(4 = (3^9)a - (2)(3^2) - (2)(3^1) - 2\)
These numbers are not huge, thus:
\(4 = 3^9a - 18 - 6 - 2\)
\(4 + 26 = 3^9a\)
\(30 = 3^9a\)
\(\frac{3^1*10^1}{3^9}= a\)
\(\frac{10}{3^{(9-1)}}= \frac{10}{3^8}= a\)
Answer B
**
For the terms whose denominators are not "canceled" by \(3^9\):
\(\frac{a^{n}}{a^{m}} = a^{(n-m)}\)
\((3^9)*(\frac{2}{3^7})=\frac{2*3^9}{3^7}= (2)(3^{(9-7)})= (2)(3^2)\) AND
\((3^9)*(\frac{2}{3^8})=\frac{2*3^9}{3^8}= (2)(3^{(9-8)})= (2)(3^1)\)can you please explain this part \(\frac{3^1*10^1}{3^9}= a\) why did you write 30 as \(3^1\) * \(10^1\) ?