logx+2^27=3
x=1

- anonymous

logx+2^27=3
x=1

- katieb

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- anonymous

\[\sqrt[3]{(x+2)^3}=\sqrt[3]{27}\]
\[x+2=3\]
\[x=1\]

- anonymous

I keep getting confused as to how I handle the \[\sqrt[3]{(x)^3}\]
and the
\[\sqrt[3]{x}\]

- 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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## More answers

- anonymous

What does it mean to take the cube root?

- freckles

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

- anonymous

So you wouldn't FOIL (x+2)?
Also 27*1/3=9

- freckles

how would you foil (x+2)?

- freckles

also why write 27*1/3

- freckles

\[(..)^\frac{1}{3} \text{ is \not the same as } (..) \cdot \frac{1}{3}\]

- 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

but why are you doing that (x+2) does not have a square on it

- 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

The exact way it is written is
\[\sqrt[3]{(x+2)^3}\]

- anonymous

Ohhh because the 3's cancel each other out?

- freckles

yes the cube function cancels the cube root function

- freckles

and you get what is inside

- 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

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

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

@Prometheus777 using that above
do you think you could simplify:
\[\sqrt[5]{32}\]

- freckles

notice the root number is 5

- freckles

like for the 3rd root I was looking for 3 of a kind

- freckles

so for the 5th root I'm looking for ...

- anonymous

The fifth of a kind

- freckles

5 of a kind
and guess what 32 has how many factors of 2?

- anonymous

32 has five of a kind

- freckles

32 has 5 factors of 2 right?

- freckles

\[\sqrt[5]{32}=2 \text{ since } 2^5=32 \\ \text{ so you can now find } \log_2(32)\]

- 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

x+6=2^5 is as far as I've gotten

- freckles

should be this:
\[(x+6)^5=32\]

- freckles

and you know from my previous example I gave 32 can be written as 2^5

- freckles

so when taking 5th root of both sides
things come out so pretty :)

- anonymous

\[\log_{x+6} 36=5\]
\[\sqrt[5]{(x+6)^5}=32^5\]
\[x+6=2^5\]

- 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

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

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

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

anyways just in case you didn't want a lesson or something I will stop bothering you :p

- anonymous

@freckles You're fine man :)

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