1
$\begingroup$

So I know that (6^2)^2 would just be (6^2) twice so (6^4) would be equal. But what if 10^2^2, would that just be 10^4 or something else? have been doing logs and log base 10 to a certain number (100) squared would equal what?(excuse my misunderstanding, in algebra 2 and way behind...

  • 0
    Excuse me: (6^2)(6^2): multiplied by itself twice – 2017-02-15
  • 1
    `^` is not associative, so when you write `10^3^3`, it is unclear whether that means $(10^3)^3=10^9$ or $10^{3^3}=10^{27}$. Note that specifically for `n^2^2`, those two interpretations amount to the same thing. – 2017-02-15
  • 0
    I meant the second or: 10^27, I don't really know how to use this program thing, but it would just be order of operations then, correct? I was researching something about log solving yesterday, and they said that adding the exponent of a log to the front is a cheating way to teach(my teacher may not be the best). if the log is squared, that means 10^3^3( the way you wrote it the second time. Where the exponent is squared, not a whole quantity. Thanks for noticing my comment. I have only done around 2 percent of my homework this year so(im a little lost) – 2017-02-15
  • 0
    "adding the exponent of a log to the front" I don't know what that means - can you give an example? (Note that the example in your comment to my answer doesn't actually work.) Also, I suggest doing more than 2 percent of your homework, going forward . . . – 2017-02-15

1 Answers 1

2

You have to distinguish between $(a^b)^c$ and $a^{(b^c)}$; for example, $$(2^2)^3=4^3=64 \quad\mbox{(or, another way: $(2^2)^3=2^6=64$)}$$ but $$2^{(2^3)}=2^8=256.$$ That is, exponentiation is not associative.

Now, when people leave out the parentheses - like "$a^{b^c}$" - they mean the second version, $a^{(b^c)}$. This is because the first version, $(a^b)^c$, can always be simplified immediately (by using the rule $(a^b)^c=a^{bc}$).

That said, your examples have an unfortunate property. Remember that $2^2=2\cdot 2$; this means that $$6^{(2^2)}=6^4=6^{2\cdot 2}=(6^2)^2,$$ even though in general it matters how we put the parentheses.

  • 0
    Is the double quotation on $a^ {b^c}$ in the 6th line important? – 2017-02-15
  • 0
    Um: what do you mean by quotes? I know now that a^(bc) is different from (a^b)^c. how does that work with logs then? we just put the exponent on a log at the front, I didn't really understand. so like (log base 2 of 2)^3 would be 2^(log base 2^3) – 2017-02-15
  • 0
    @MrAP I wouldn't say it's important, but it's deliberate - I'm referring to the *string of symbols* there, rather than the number represented, so I put quotes around it (similarly to how I would write '"word" is a word'). – 2017-02-15
  • 0
    @IDKTHedude Those two expressions are *not* the same - the first equals $1$, and the second equals $3$. (Remember that $\log_aa=1$, and $1^b=1$ . . .) – 2017-02-15
  • 0
    @IDKTHedude Just checking - what does "log base 2^3" mean? Do you mean $\log_23$, or something else? – 2017-02-15
  • 0
    I meant 2^(log base 2 of 2^3) how do you get exponent writing on here? – 2017-02-15
  • 0
    @IDKTHedude Use *math mode*: write an expression with a \$ on each side. E.g. if I write "\$ 2^3 \$," I'll get "$2^3$". (Other tricks: for an exponent with more than one symbol, use braces - "\$ 2^{3+8}\$" gives me "$2^{3+8}$". For subscripts, use "_".) As to the math, notice that your first expression - $(\log_22)^3$ - is just $1^3=1$, while $2^{\log_2(2^3)}=2^3=8$, so they're not equal. – 2017-02-15
  • 0
    don't know how logs would work but thanks. Do i seem way behind for Algebra 2? i also dont know about mulitplying and dividing rational expressions: he usually just writes a very complicated polynomial over polynomial, crosses out things that are in common, then gets the right answer. Why is it that you can multiply the numerator and denominator and have the same proportion, but can't use that with say axb. I will reword, axb=c you can have b/c and as long as you multiply them by the same thing, it is proportional, but not on mulitplication, or inverse function. is factoring the same thing – 2017-02-15
  • 0
    @IDKTHedude "Do i seem way behind for Algebra 2?" I couldn't say. But it does sound like you are unfamiliar with the concepts your class is using. I think you should talk to your teach about what you're having trouble with, and your level of preparation for the class. Meanwhile, I don't really understand the rest of your comment, but this may help: when you multiply the numerator and denominator by the same thing, what you're really doing is *multiplying the whole thing by $1$*: $${ac\over bc}={a\over b}\cdot {c\over c}={a\over b}\cdot 1.$$ You can multiply anything by $1$ without changing it. – 2017-02-16