Hard Question
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Otis
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Hi. I didn't see anything to try with logarithms, so I'd looked for a pattern by trying smaller numbers.
(1 + 2)(1^2 + 2^2) = 2^x - 1^x
Can you work that out by trial and error? (It helps to be familiar with beginning powers of 2.)
Then try this one.
(1 + 2)(1^2 + 2^2)(1^4 + 2^4) = 2^x - 1^x
Do you see a pattern forming?
(1 + 2)(1^2 + 2^2) = 2^x - 1^x
Can you work that out by trial and error? (It helps to be familiar with beginning powers of 2.)
Then try this one.
(1 + 2)(1^2 + 2^2)(1^4 + 2^4) = 2^x - 1^x
Do you see a pattern forming?
Cubist
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I'll give an extra hint, since it's a day later...
[imath](6+5) (6^2+5^2) \cdots[/imath]
[imath]= \color{red}1\times\color{black}(6+5) (6^2+5^2)\cdots[/imath]
[imath]= \color{red}(6-5) \color{black}(6+5) (6^2+5^2)\cdots[/imath]
[imath]= (6^2-5^2) (6^2+5^2) \cdots[/imath]
[imath]= (6^4-5^4)\cdots[/imath]
Hi. I didn't see anything to try with logarithms, so I'd looked for a pattern by trying smaller numbers.
(1 + 2)(1^2 + 2^2) = 2^x - 1^x
Can you work that out by trial and error? (It helps to be familiar with beginning powers of 2.)
Then try this one.
(1 + 2)(1^2 + 2^2)(1^4 + 2^4) = 2^x - 1^x
Do you see a pattern forming?
Thank you both for the responses! I was able to find x but I think there may be a gap in my knowledge/understanding because I don't fully understand the working out. I would really appreciate a more detailed explanation if possible? But thank you for taking the time to respond to my question!
Otis
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Hi. Please show us what you did to find the x-solution. Can you identify for us the part(s) of your work for which you sense a lack of understanding? Thanks!
The way I comprehended it, I saw that x would always equal double the exponent in the last bracket. e.g: if it was (6^2 + 5^2), x = 4; if it was (6^32 + 5^32), x = 64
And I'm not sure if I was supposed to get the answer like that? ?
With @Cubist 's response, I didn't really understand how a difference of squares was involved and with your response, I wasn't exactly sure if I was following it correctly.
Thanks again
And I'm not sure if I was supposed to get the answer like that? ?
With @Cubist 's response, I didn't really understand how a difference of squares was involved and with your response, I wasn't exactly sure if I was following it correctly.
Thanks again
Otis
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How did you see it? A guess or calculations.
You haven't posted any context or instructions that apply, so we're free to solve such problems by inspection, educated guesses and so on.
Well, did you try it? That is, did you multiply out the first few factors, to see a pattern? If you need to turn in a written answer, that would be a nice demonstration based on solid algebra.
Maybe this will help you. A difference of even powers like a^4 - b^4 may be viewed as a difference of squares. We can see it as (a^2)^2 - (b^2)^2 so it factors as (a^2 + b^2) (a^2 - b^2).
Reason through it on paper, starting with what you know. You can multiply binomials, yes? That is, double distribution, also known by FOIL. Do the multiplications shown in post #4. Extend the pattern.