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2 ^ 28, but how was this done without using a calculator. The furthest I can take this without using a calculator is 2 ^ 29 - 2 ^ 28

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- #1

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2 ^ 28, but how was this done without using a calculator. The furthest I can take this without using a calculator is 2 ^ 29 - 2 ^ 28

- #2

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2 ^ 29 - 2 ^ 28 = 2^28(2-1) = 2^28(1) = 2^28

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HallsofIvy

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(2-1)2

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- #5

Gokul43201

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Complete solutions have been posted above. What specific help are you looking for? If you explain which concept you're having difficulty with, someone might be able to help you better.

Edit : If your problem is entirely unrelated to the prior content of this thread, state the COMPLETE question in a new thread.

Edit : If your problem is entirely unrelated to the prior content of this thread, state the COMPLETE question in a new thread.

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- #7

Office_Shredder

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(2

Step one: cancel the two in the denominator. TO do this, you divide each term by 2, meaning (for this specific scenario), you subtract one from each exponent

2

Step two: Factor 2

2

Step three: Remembering some basic exponent rules, we clean up the equation a bit (specifically a number to the power of 1 or 0)

2

Step four: Finish solving by calculating 2-1

2

- #8

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3^(n+2) / [3^(n+3) - 3^(n+1)]

the answer key says that the answer is 1/3 but when I solved for it, the answer is 3/8.. what is really the correct answer and how will it be solved?

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HallsofIvy

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The correct answer is 1/3 as you were told- write [itex]3^{n+2}[/itex] as [itex]3(3^{n+1})[/itex] and cancel.

- #10

Mute

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Unless you happen to have [itex]8 = 3^2-3^0[/itex].If you only have powers of 3, you cannot possibly have a factor of [itex]8= 2^3[/itex]!

I get 3/8:The correct answer is 1/3 as you were told- write [itex]3^{n+2}[/itex] as [itex]3(3^{n+1})[/itex] and cancel.

[tex]\frac{3^{n+2}}{3^{n+3}-3^{n+1}} = \frac{3 3^{n+1}}{3^2 3^{n+1}-3^{n+1}} = \frac{3}{3^2-3^0} = \frac{3}{9-1} = \frac{3}{8}[/tex]

Perhaps HallsofIvy and the person who wrote the book's answer key missed the "-3^(n+1)" in the denominator?

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