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# If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?

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If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?  [#permalink]

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11 Jan 2018, 23:18
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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

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If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?  [#permalink]

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13 Jan 2018, 12:29
1
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

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
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Re: If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?  [#permalink]

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20 Jan 2018, 20:12
1
1
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:

$$4 = 3^9a - 18 - 6 - 2$$

$$4 + 26 = 3^9a$$

$$30 = 3^9a$$

$$\frac{3^1*10^1}{3^9}= a$$

$$\frac{10}{3^{(9-1)}}= \frac{10}{3^8}= a$$

**
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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If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?  [#permalink]

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21 Jan 2018, 02:15
generis wrote:
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:

$$4 = 3^9a - 18 - 6 - 2$$

$$4 + 26 = 3^9a$$

$$30 = 3^9a$$

$$\frac{3^1*10^1}{3^9}= a$$

$$\frac{10}{3^{(9-1)}}= \frac{10}{3^8}= a$$

**
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

$$(3^9)*(\frac{2}{3^8})=\frac{2*3^9}{3^8}= (2)(3^{(9-8)})= (2)(3^1)$$

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
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If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?  [#permalink]

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21 Jan 2018, 08:45
1
dave13 wrote:
generis wrote:
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

$$\frac{4}{3^9} = a - \frac{2}{3^7} - \frac{2}{3^8} - \frac{2}{3^9}$$

**Multiply all terms by $$3^9$$ =

$$4 = (3^9)a - (2)(3^2) - (2)(3^1) - 2$$

These numbers are not huge, thus:

$$4 = 3^9a - 18 - 6 - 2$$

$$4 + 26 = 3^9a$$

$$30 = 3^9a$$

$$\frac{3^1*10^1}{3^9}= a$$

$$\frac{10}{3^{(9-1)}}= \frac{10}{3^8}= a$$

**
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

$$(3^9)*(\frac{2}{3^8})=\frac{2*3^9}{3^8}= (2)(3^{(9-8)})= (2)(3^1)$$

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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Re: If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?  [#permalink]

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21 Jan 2018, 09:38
1
1
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

$$\frac{4}{3^9} = a - \frac{2}{3^7} - \frac{2}{3^8} - \frac{2}{3^9}$$

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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Re: If 4/3^9 = a - 2/3^7 - 2/3^8 - 2/3^9, then what is the value of a?   [#permalink] 21 Jan 2018, 09:38
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