# How do you solve 3^n=81?

Asked by Kitanrum (4) August 7th, 2010

I know the answer is 4, but I don’t understand how it’s done. Can any one help me? =(

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I don’t want to totally give it away, but you need to use logarithms..

cockswain (15271)

That would have helped… If I knew logarithms. I need like a step by step answer. Not giving an answer doesn’t help my cause.

Kitanrum (4)

@Kitanrum you see, the thing is, this sounds very much like a homework question (not saying it is, just saying it sounds like one) and we don’t do homework questions. We’ll give you a hand and point you in the right direction, but thats about it.

Lightlyseared (32243)

@Lightlyseared How old are you and what grade are you in? That would help us figure out what level math you are at.

Rarebear (25159)

@Lightlyseared Well it’s not a homework question. It is from a textbook, a British one. I’m reviewing for a math placement exam for college so that I can be placed in a good math class without being considered a half-time student on the count that I could lose scholarship money. That’s the only reason I’m asking. So if I could get an answer, that would be great.

Kitanrum (4)

This will be helpful. It’s an easy to understand tutorial on what you’re trying to learn. Let me know if you have more questions.

cockswain (15271)

@cockswain Thaaaaaank you!!! >_< Learning math on my own is kinda hard and that makes it super easy! Thank you!!!!

Kitanrum (4)

3 ^ n = 81
log 3 ^ n = log 81 (log both side)
n log 3 = log 81 (log 3^n is the same as n log 3)
n = log 81 / log 3 (divide by log 3)

solve on a calculator

Ame_Evil (3046)

@Ame_Evil D: That’s even better! Thank you so much!

Kitanrum (4)

You can solve this without a calculator. 81= 9*9 = (3*3)*(3*3) = 3^4.

@LostInParadise True, but that doesn’t really help the OP understand the underlying concept. I mean, I memorized my squares and cubes a long time ago, but it wasn’t until a few years after that (my high school years) that I understood logarithms.

jerv (31044)

@Kitanrum: I strongly advise you to learn the concept of logarithms (at least to the base 10).

It is simple and should be one chapter in a HS math textbook.

gailcalled (54577)

@jerv, First the student has to get an intuitive feel for how exponents work. Then it is possible to understand the more advanced material on how to manipulate logarithms.

@LostInParadise Precisely! And it takes more than memorizing tables to get to that point.

jerv (31044)

@Kitanrum, do you at least understand that 3 to the 4th power (which we write as 3^4) equals 81 ?
Or is that the part you don’t get?

gasman (11313)

@gasman, good point. Not to sidetrack this discussion, but I think exponential notation is unnecessarily confusing. If we understand 4*3 to mean 3+3+3+3, wouldn’t it make more sense to use 4^3 to mean 3*3*3*3? In this case we know exactly who to blame. Our exponential notation is due to Descartes, who would seem (oh please forgive me) to have put de cart before the horse.

@LostInParadise My calculator of choice is an old HP 48G that uses Reverse Polish Notation. I often bet people that they can’t add 2 and 2 on my calculator and often win; they hit 2 and then the + key, which trips a Syntax error. (The correct response is “2 ENTER 2 +”)

jerv (31044)

@LostInParadise 4*3 also means 4+4+4—it just so happens that it’s a commutative operation and your statement is true as well. The notation is not really inconsistent, just a choice that was made. Exponentiation is not commutative, so it can only work one way.

lapilofu (4325)

This will have to be my last comment on this side discussion, not because it is not interesting (it deserves its own thread) but because I do not want to hijack the original question.

@jerv
With regard to Polish notation, I find that once you get the hang of it, it can be less confusing than entering parentheses. (2+3) * (4+7) can be a bit confusing to enter sequentially, especially if you don’t have a graphing calculator, whereas 2 3+4 7+* is not as bad.

@lapilofu
Technically speaking, 4*3 is not the same as 3*4. The second term is called the multiplicand and it is what gets multiplied by the first term, called the multiplier. I admit that this is a little pedantic, since both expressions have the same value and the distinctions break down when you get beyond integers. Still, it does show that there is an inconsistency with exponential notation, where, as you point out, order makes a difference since the operation is not commutative.

@LostInParadise My sentiments exactly! I have a hard time with the limitations of a “normal” calculator now.

jerv (31044)

@LostInParadise Agreed that notation is awful, but without superscripts we’re stuck with typographical conventions of the internet. There’s no real substitute (is there?) for an expression like (x^y). Besides, replacing 3^4 by 3*3*3*3 won’t work for non-integer exponents and misses the point of introducing the concept of logarithms.

@jerv My first (and best !) calculator was an HP-25. w/ RPN. Later I did some programming the language FORTH, which works in a similar fashion.

I, too, pledge no more off-topic stuff here. My apologies, @Kitanrum.

gasman (11313)

\3/^81=4

@LostInParadise Well perhaps it’s multiplication then which has the inconsistent notation? In exponentiation (a^b) a is the quantity which you raise to b. In division (a/b) a is the quantity (the dividend) that is divided by b (divisor). Why shouldn’t it be in multiplication (a*b) that a is the quantity (multiplicand) you multiply by b (multiplier)? What is the reason for the multiplier coming first?

lapilofu (4325)

Addition and multiplication are said to be commutative operations, because a ~ b = b ~ a where the tilde stands for some operation between two numbers a and b. This is another way of saying (for multiplication) that three sevens equals seven threes, etc, or (for addition) that two more than nine equals nine more than two.

Subtraction, division, and exponentiation are non-commutative—the order matters. In some mathematical systems (matrices, for example) a x b does not necessarily equal b x a. We say that a is multiplied by b (in some well-defined way) so the result is not ambiguous.

gasman (11313)

When mathematics goes beyond syntax and into outright semantics, you are entering into a realm where many (myself included) just roll their eyes, walk away, and try to get drunk enough to forget the whole thing.

I am looking at my own brain and walking away now….

jerv (31044)

Well, Ame_Evil give the best algorithm. I almost forget this algorithm.

Ailsa01 (16)

or