anonymous
  • anonymous
logx+2^27=3 x=1
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\sqrt[3]{(x+2)^3}=\sqrt[3]{27}\] \[x+2=3\] \[x=1\]
anonymous
  • anonymous
I keep getting confused as to how I handle the \[\sqrt[3]{(x)^3}\] and the \[\sqrt[3]{x}\]
freckles
  • freckles
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: \[\sqrt[3]{u^3}=u\]

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anonymous
  • anonymous
What does it mean to take the cube root?
freckles
  • freckles
\[\text{ this cube \root function } \rightarrow \sqrt[3]{ .. } \text{ or \it can also be written as } (..)^\frac{1}{3}\]
anonymous
  • anonymous
So you wouldn't FOIL (x+2)? Also 27*1/3=9
freckles
  • freckles
how would you foil (x+2)?
freckles
  • freckles
also why write 27*1/3
freckles
  • freckles
\[(..)^\frac{1}{3} \text{ is \not the same as } (..) \cdot \frac{1}{3}\]
anonymous
  • anonymous
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)
freckles
  • freckles
but why are you doing that (x+2) does not have a square on it
freckles
  • freckles
remember when I said \[\sqrt[3]{u^3}=u \\ \text{ so that means }\sqrt[3]{(x+2)^3}=x+2 \text{ or } \sqrt[3]{3^3}=3\]
anonymous
  • anonymous
The exact way it is written is \[\sqrt[3]{(x+2)^3}\]
anonymous
  • anonymous
Ohhh because the 3's cancel each other out?
freckles
  • freckles
yes the cube function cancels the cube root function
freckles
  • freckles
and you get what is inside
freckles
  • freckles
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: \[5^3 \text{ means } 5 \cdot 5 \cdot 5 =125 \\ \text{ and if we do the cube \root of 125...} \\ \sqrt[3]{125}=5 \text{ since } 5^3=125 \\ \text{ basically} \text{ if } x^3=a \text{ then } \sqrt[3]{a}=x \\ \text{ more examples } \\ 2^3=8 \text{ so } \sqrt[3]{8}=2 \\ (-1)^3=-1 \text{ so } \sqrt[3]{-1}=-1 \\ 4^3=64 \text{ so } \sqrt[3]{64}=4 \\ 10^3=1000 \text{ so } \sqrt[3]{1000}=10\]
anonymous
  • anonymous
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 :)
freckles
  • freckles
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
freckles
  • freckles
@Prometheus777 using that above do you think you could simplify: \[\sqrt[5]{32}\]
freckles
  • freckles
notice the root number is 5
freckles
  • freckles
like for the 3rd root I was looking for 3 of a kind
freckles
  • freckles
so for the 5th root I'm looking for ...
anonymous
  • anonymous
The fifth of a kind
freckles
  • freckles
5 of a kind and guess what 32 has how many factors of 2?
anonymous
  • anonymous
32 has five of a kind
freckles
  • freckles
32 has 5 factors of 2 right?
freckles
  • freckles
\[\sqrt[5]{32}=2 \text{ since } 2^5=32 \\ \text{ so you can now find } \log_2(32)\]
freckles
  • freckles
well actually you could probably have found that without know the 5 root part :) but let's see if you can solve this: \[\log_{x+6}(32)=5\]
anonymous
  • anonymous
x+6=2^5 is as far as I've gotten
freckles
  • freckles
should be this: \[(x+6)^5=32\]
freckles
  • freckles
and you know from my previous example I gave 32 can be written as 2^5
freckles
  • freckles
so when taking 5th root of both sides things come out so pretty :)
anonymous
  • anonymous
\[\log_{x+6} 36=5\] \[\sqrt[5]{(x+6)^5}=32^5\] \[x+6=2^5\]
freckles
  • freckles
\[\log_a(x)=y \text{ is equivalent to }a^y=x \\ \log_{x+6}(32)=5 \text{ is equivalent to } (x+6)^5=32\] \[(x+6)^5=32 \\ \text{ now recall that } 2^5=32 \text{ so we have: } \\ (x+6)^5=2^5 \text{ now taking 5th root of both sides } \\ \sqrt[5]{(x+6)^5}=\sqrt[5]{2^5} \\ \text{ I said this would become pretty because everything we had } \\ \text{ we were able to write the insides as perfect powers of 5 } \\ \text{ so the 5 root will cancel the power of 5 } \\ (x+6)^\frac{5}{5}=2^\frac{5}{5} \\ (x+6)=2 \text{ since the 5th root canceled the power of 5 } \\ \text{ or you can also look at it as well } \frac{5}{5}=1 \\ \text{ so now you have the equation } x+6=2\]
freckles
  • freckles
we actually already knew from earlier that: \[2^5=32 \text{ which means } \log_2(32)=5 \ \\ \text{ and our equation was } \\ \log_{x+6}(32)=5 \text{ upon comparing this to } \\ \log_2(32)=5 \\ \text{ we should actually be able to say without all this work that } \\ x+6=2 \text{ from comparing those two equations }\]
freckles
  • freckles
The problem you had earlier: The: \[\log_{x+2}(27)=3 \\\ \text{ we know } \log_3(27)=3 \text{ since } 3^3=27 \\ \text{ so comparing } \\ \log_{x+2}(27)=3 \text{ to } \log_3(27)=3 \\ \text{ without all the work we did above we should be able to say } x+2=3\]
freckles
  • freckles
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.
freckles
  • freckles
anyways just in case you didn't want a lesson or something I will stop bothering you :p
anonymous
  • anonymous
@freckles You're fine man :)

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