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anonymous

  • one year ago

logx+2^27=3 x=1

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  1. anonymous
    • one year ago
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    \[\sqrt[3]{(x+2)^3}=\sqrt[3]{27}\] \[x+2=3\] \[x=1\]

  2. anonymous
    • one year ago
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    I keep getting confused as to how I handle the \[\sqrt[3]{(x)^3}\] and the \[\sqrt[3]{x}\]

  3. freckles
    • one year ago
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    I guess the oringal equation was: \[\log_{x+2}(27)=3\] since your equivalent exponential you wrote was \[(x+2)^3=27\] and you can write 27 as 3^3 \[(x+2)^3=3^3\] and you just take cube root of both sides and we know that for any number u we have: \[\sqrt[3]{u^3}=u\]

  4. anonymous
    • one year ago
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    What does it mean to take the cube root?

  5. freckles
    • one year ago
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    \[\text{ this cube \root function } \rightarrow \sqrt[3]{ .. } \text{ or \it can also be written as } (..)^\frac{1}{3}\]

  6. anonymous
    • one year ago
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    So you wouldn't FOIL (x+2)? Also 27*1/3=9

  7. freckles
    • one year ago
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    how would you foil (x+2)?

  8. freckles
    • one year ago
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    also why write 27*1/3

  9. freckles
    • one year ago
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    \[(..)^\frac{1}{3} \text{ is \not the same as } (..) \cdot \frac{1}{3}\]

  10. anonymous
    • one year ago
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    I see my mistake in the second area For the foiling take the product of (x+2)(x+2) and foil that once more against (x+2)

  11. freckles
    • one year ago
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    but why are you doing that (x+2) does not have a square on it

  12. freckles
    • one year ago
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    remember when I said \[\sqrt[3]{u^3}=u \\ \text{ so that means }\sqrt[3]{(x+2)^3}=x+2 \text{ or } \sqrt[3]{3^3}=3\]

  13. anonymous
    • one year ago
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    The exact way it is written is \[\sqrt[3]{(x+2)^3}\]

  14. anonymous
    • one year ago
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    Ohhh because the 3's cancel each other out?

  15. freckles
    • one year ago
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    yes the cube function cancels the cube root function

  16. freckles
    • one year ago
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    and you get what is inside

  17. freckles
    • one year ago
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    I think you don't know what cube root means so maybe look at this and see if this better clears up what it means: Examples of cube root function (along with cube function): \[\text{cube function } (.. )^3 \\ \\ \text{ cube \root function } \sqrt[3]{ ..} \text{ or } (..)^\frac{1}{3}\] examples: \[5^3 \text{ means } 5 \cdot 5 \cdot 5 =125 \\ \text{ and if we do the cube \root of 125...} \\ \sqrt[3]{125}=5 \text{ since } 5^3=125 \\ \text{ basically} \text{ if } x^3=a \text{ then } \sqrt[3]{a}=x \\ \text{ more examples } \\ 2^3=8 \text{ so } \sqrt[3]{8}=2 \\ (-1)^3=-1 \text{ so } \sqrt[3]{-1}=-1 \\ 4^3=64 \text{ so } \sqrt[3]{64}=4 \\ 10^3=1000 \text{ so } \sqrt[3]{1000}=10\]

  18. anonymous
    • one year ago
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    That makes so much sense now, Thank you for your patience and Thank you for walking me through the logic that is the cube root and cube root function :)

  19. freckles
    • one year ago
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    sometimes the number inside won't be a perfect cube but you can still rewrite if the factors contained in the number inside are perfect cubes for example: \[\sqrt[3]{32} \\ \text{ it may help to write the prime factorization for } 32 \\ 32=2^5 \\ \text{ now are are looking for factors that come in triplets } \\ 32=2 \cdot 2 \cdot 2 \cdot 2 \ \cdot 2 \\ \text{ I see a triplet } \\ 32=(2 \cdot 2 \cdot 2) \cdot 2 \cdot 2 \\ 32=2^3 \cdot 2 \cdot 2 \\ 32 = 2^3 \cdot 4\] \[\sqrt[3]{32}=\sqrt[3]{2^3 \cdot 4} =\sqrt[3]{2^3} \cdot \sqrt[3]{4}=2 \sqrt[3]{4} \text{ since } \sqrt[3]{u^3}=u \\ \text{ so we have } \sqrt[3]{32} \text{ can be expressed as } 2 \cdot \sqrt[3]{4} \text{ which is more commonly written as } \\ 2 \sqrt[3]{4} \text{ but is still the same }\] anyways this is just a little more advanced than the previous examples and you probably won't need it for these log problems but it is good to know just in case you are asked to rewrite a cube root expression in this way I still think it is a good example for anyone tying to get the hang of cube roots

  20. freckles
    • one year ago
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    @Prometheus777 using that above do you think you could simplify: \[\sqrt[5]{32}\]

  21. freckles
    • one year ago
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    notice the root number is 5

  22. freckles
    • one year ago
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    like for the 3rd root I was looking for 3 of a kind

  23. freckles
    • one year ago
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    so for the 5th root I'm looking for ...

  24. anonymous
    • one year ago
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    The fifth of a kind

  25. freckles
    • one year ago
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    5 of a kind and guess what 32 has how many factors of 2?

  26. anonymous
    • one year ago
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    32 has five of a kind

  27. freckles
    • one year ago
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    32 has 5 factors of 2 right?

  28. freckles
    • one year ago
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    \[\sqrt[5]{32}=2 \text{ since } 2^5=32 \\ \text{ so you can now find } \log_2(32)\]

  29. freckles
    • one year ago
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    well actually you could probably have found that without know the 5 root part :) but let's see if you can solve this: \[\log_{x+6}(32)=5\]

  30. anonymous
    • one year ago
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    x+6=2^5 is as far as I've gotten

  31. freckles
    • one year ago
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    should be this: \[(x+6)^5=32\]

  32. freckles
    • one year ago
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    and you know from my previous example I gave 32 can be written as 2^5

  33. freckles
    • one year ago
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    so when taking 5th root of both sides things come out so pretty :)

  34. anonymous
    • one year ago
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    \[\log_{x+6} 36=5\] \[\sqrt[5]{(x+6)^5}=32^5\] \[x+6=2^5\]

  35. freckles
    • one year ago
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    \[\log_a(x)=y \text{ is equivalent to }a^y=x \\ \log_{x+6}(32)=5 \text{ is equivalent to } (x+6)^5=32\] \[(x+6)^5=32 \\ \text{ now recall that } 2^5=32 \text{ so we have: } \\ (x+6)^5=2^5 \text{ now taking 5th root of both sides } \\ \sqrt[5]{(x+6)^5}=\sqrt[5]{2^5} \\ \text{ I said this would become pretty because everything we had } \\ \text{ we were able to write the insides as perfect powers of 5 } \\ \text{ so the 5 root will cancel the power of 5 } \\ (x+6)^\frac{5}{5}=2^\frac{5}{5} \\ (x+6)=2 \text{ since the 5th root canceled the power of 5 } \\ \text{ or you can also look at it as well } \frac{5}{5}=1 \\ \text{ so now you have the equation } x+6=2\]

  36. freckles
    • one year ago
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    we actually already knew from earlier that: \[2^5=32 \text{ which means } \log_2(32)=5 \ \\ \text{ and our equation was } \\ \log_{x+6}(32)=5 \text{ upon comparing this to } \\ \log_2(32)=5 \\ \text{ we should actually be able to say without all this work that } \\ x+6=2 \text{ from comparing those two equations }\]

  37. freckles
    • one year ago
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    The problem you had earlier: The: \[\log_{x+2}(27)=3 \\\ \text{ we know } \log_3(27)=3 \text{ since } 3^3=27 \\ \text{ so comparing } \\ \log_{x+2}(27)=3 \text{ to } \log_3(27)=3 \\ \text{ without all the work we did above we should be able to say } x+2=3\]

  38. freckles
    • one year ago
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    I don't know if this makes these type of problems easier for you to solve or not but it is probably best to know both ways because your equations will not always come out with pretty solutions like yours have been for example: \[\log_{x}(3)=5 \\ \text{ is equivalent to } x^5=3 \\ \text{ take 5th root of both sides } \sqrt[5]{x^5}=\sqrt[5]{3} \\ \text{ so we have the solution is } x=\sqrt[5]{3}\] This solution is not as nice looking at the ones before.

  39. freckles
    • one year ago
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    anyways just in case you didn't want a lesson or something I will stop bothering you :p

  40. anonymous
    • one year ago
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    @freckles You're fine man :)

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