# How come $2 \times 3^k + 3^k = 3 \times 3^k$

I get something else

\begin{align*} &2 \times 3^k + 3^k=\\ &2 \times 3^k \times 2=\\ &4 \times 3^k \end{align*}

What does $3^k + 3^k$ give exactly?

-
No matter what $a$ is, $2a+a=3a$; now just let $a=3^k$. –  Brian M. Scott Dec 5 '12 at 23:18

Remember the order of precedence of operations. Here,

First Multiply $\quad2\times 3^k\quad$ ... then add: $\quad(2\times 3^k) + 3^k\;$: $$2\times 3^k + 3^k = (2 \times 3^k) + 3^k = (3^k + 3^k) + 3^k = 3\times 3^k$$

Note that if we just let $x = 3^k$, then $$2 \times 3^k + 3^k = (2\times x) + x = (2 \times x) + (1\times x) = (2 + 1)\times x = 3\times x.$$

Now, since we let $x = 3^k$, then $3 \times x = 3\times 3^k$.

For your last question, $3^k + 3^k = 2\times 3^k$, but note that the first operation to perform in your original question is $\quad(1st)$ "$\times$" $\quad$ then $\;\;(2nd)$ "$+$".

-
+100 for my dear friend Amy. ;^) –  Babak S. Mar 27 '13 at 7:15

Your calculation appears to be based on the notion that $2\times 3^k+3^k$ means $2\times(3^k+3^k)$, which would indeed be $4\cdot3^k$. However, in the absence of parentheses multiplication is performed before addition, so $2\times 3^k+3^k$ actually means $(2\times 3^k)+3^k$. This is $(2\times 3^k)+(1\times 3^k)$, which is clearly just $3\times 3^k$, or $3^{k+1}$.

-

It looks like you are confusing $2\cdot 3^k+3^k$ with $2 \cdot (3^k+3^k)$. The first gives $3 \cdot 3^k=3^{k+1}$, while the second gives $2 \cdot (2 \cdot 3^k)=4\cdot 3^k$. Parentheses are important.

-

$2\times 3^k+3^k=(2+1)\times3^k=3\times3^k$

-