Why is $3 \cdot 3^k = 3^{k+1}$ and not $9^k$?

Question

Why is $3 \cdot 3^k = 3^{k+1}$ and not $9^k\;$?

I'm aware that $3 = 3^1$ but I would expect $3\cdot 3^k\;$ to be $\;9^k$ or $\;9^{k+1}$.

Answer 1

$$3\cdot 3^k = 3\cdot\underbrace{3\cdot 3 \cdot \cdots 3 \cdot 3}_{k\;factors \;of \;3}$$ $$= \underbrace{3\cdot 3 \cdot \cdots 3 \cdot 3}_{k+1\;factors \;of \;3}$$ $$= 3^{k+1}$$

By convention, taking powers (exponentiation) precedes multiplication, and so it is performed first (before multiplication). That is, think of $\;3\cdot 3^k\;$ as expressing $\;3(3^k) = 3^{k+1}$.

If multiplication is to precede exponentiation, one would need to use parentheses to indicate "multiply first, then take the exponent of that product": $(3\cdot 3)^k = 9^k$.

$$\text{But}\;\; 3\cdot (3^k) \neq (3\cdot 3)^k$$ $$3^{k+1} \neq 9^k \text{ unless k = 1}. $$

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If you really like the number $9$ and really want to use it in expressing $3\cdot3^k$, then note that $$3\cdot 3^k = 3\cdot\underbrace{3\cdot 3 \cdot \cdots 3 \cdot 3}_{k \;factors \;of \;3} = 9\cdot 3^{k-1} = 3\cdot 3 \cdot \underbrace{3\cdot 3 \cdot \cdots 3 \cdot 3}_{(k-1) \;factors \;of \;3}$$

Answer 2

The answer is as follows: Because there's a convention that powers have larger priority than multiplication, therefore $$ 3\cdot 3^k = 3\cdot(3^k) \qquad\text{ and not }\qquad 3\cdot 3 ^k = (3\cdot 3)^k.$$

The question is similar to asking why $3+4\cdot 5=3+20=23$ and not $3+4\cdot5=7\cdot5=35$.

Answer 3

From my point of view, The Answer is: It just does not follow law of indices, and nothing more

Answer 4

We have $a^b\cdot a^c=a^{b+c}$. Look at this: $$\underbrace{a\cdot a\cdots \cdot a}_{b\text{ factors}}\cdot \underbrace{a\cdot a\cdots \cdot a}_{c\text{ factors}}=\underbrace{a\cdot a\cdots \cdot a}_{b+c\text{ factors}}$$

Answer 5

The left-hand side of your equation involves two operations, multiplication and raising to a power. The agreed-upon convention for such expressions is to perform the exponentiation first (we assign it a higher precedence) and then do the multiplication. Using parentheses to force this operation we'd have, for $k=2$, for example $$ 3\cdot3^2=3\cdot(3^2)=3\cdot(9)=27 $$ Doing the multiplication first we'd have $$ 3\cdot3^2=(3\cdot3)^2=(9)^2=81 $$ Mathematicians could have agreed that the order would be "multiplication first, and then exponentiation", but they didn't: absent parentheses, exponentiation (and other functions like $\log,\sin$ and negation) are performed first, then any multiplications and divisions are performed, and finally any additions and subtractions.

Answer 6

By convention $\rm\: a\cdot b^{\,n}\:$ means $\rm\:a \cdot (b^n)\:$ not $\rm\:(a\cdot b)^n.\:$ If we adopted the alternative convention, then many common arithmetical expressions (e.g. polynomials) would require more symbols to notate.