Can't use a calculator
\[\frac{2 ^{30} -2^{29}}{ 2 }\]
Can I reduce it? Like 2^4-2^3 / 2 and it turn out the same?
Factor out 2^29 from the numerator.
Hum. Can you show me how to do that with the exp? I understand factoring but I'm afraid I will kill it.
Example: \(x^{30} - x^{29} = x^{29}(x - 1)\) Now replace x with 2.
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Suppose you have `n` apples. If you throw away half of them, how many apples you will be left with ?
*Without a calculator* Ok @mathstudent55 \[\frac{ 2^{29}(x-1) }{ 2 }\]
Replace x with 2 in the parentheses also.
Oh yea my bad
\[\frac{ 2^{29}(2-1) }{ 2 }\]
Now what is 2 - 1? It's just numbers, so you can do the subtraction.
1
So just 2^29 on top
Right, so you have \(\dfrac{2^{29}}{2^1} \) How do you divide powers with the same base?
Subtract, right?
Correct.
So I am at \[2^{28}\]
That is correct.
Ok. Well, thankfully I can stop there...it does not want an actual number value. Thanks!
Now just give me a minute. Look above at what @ganeshie8 wrote.
Notice that when you have powers of 2, each next higher integer power is 2 times the previous power. This is what I mean. For example, \(2^{30} = 2^{29} \times 2\) \(2^{15} = 2^{14} \times 2\) In general, \(2^{n + 1} = 2^n \times 2\) Ok so far?
Yes, I follow.
And thanks for taking so much time to explain this BTW.
That means each power of 2 is half of the next higher integer power of 2. In general \(2^n = \dfrac{2^{n + 1}}{2} \) That means, for the powers used in this problem, \(2^{29} = \dfrac{2^{30}}{2} \)
Wow thanks so much!
That means in the numerator of the fraction, where you have \(2^{30} - 2^{29}\), you are subtracting half of \(2^{30}\). Then you can conclude immediately that \(2^{30} - 2^{29} = 2^{29} \)
Then when you divide \(2^{29}\) by 2, as the denominator of the fraction shows, you get \(2^{28} \)
This second approach is all based on what @ganeshie8 asked above.
You're welcome.
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