For simplicity let \(\frac{2}{3^9}=x\) so our equation becomes

\(2x=a-3^2x-3x-x=a-9x-3x-x\)

\(=>a=15x=15*\frac{2}{3^9} = \frac{10}{3^8}\)

Option B

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

\((3^9)*(\frac{2}{3^8})=\frac{2*3^9}{3^8}= (2)(3^{(9-8)})= (2)(3^1)\)
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Hi generis,

can you please explain this part \(\frac{3^1*10^1}{3^9}= a\) why did you write 30 as \(3^1\) * \(10^1\) ?

and this part \(\frac{10}{3^{(9-1)}}= \frac{10}{3^8}= a\)

why did you subtract 1 from 9 in the denominators exponent ?

thanks and have a great day

dave13 , I factored 30 because no answer choice has 30 as a numerator. Three choices have 10 as a numerator. I needed to get rid of the 3. (I wrote the exponents so that my next step would be clear.)

Second, I could have written \(3^{9+(-1)}\). Same thing.
When we have the same base (3), with different exponents (1 and 9), and we are dividing or multiplying, what do we do?(Hint: see my footnote)

Please take a look at Bunuel EXPONENTS

Purplemath,Exponents - Basic Rules, and Simplifying Expressions
Math Planet,Powers and Exponents

Hope that helps.
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\(\frac{4}{3^9} = a - \frac{2}{3^7} - \frac{2}{3^8} - \frac{2}{3^9}\)

Or, \(4 = a*3^9- 2*3^2 - 2*3 - 2\)

Or, \(4 = a*3^9 - 18 - 6 - 2\)

Or, \(a = \frac{30}{3^9}\)

Or, \(a = \frac{10}{3^8}\) , Answer will hence be (B)
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