Simplifying $(m^3)^4(2x^3)^7(m^2)^5(3x)^2$

Question

I am having trouble with simplifying the following algebraic expression:

$$(m^3)^4(2x^3)^7(m^2)^5(3x)^2$$

I have been able to do the exponents and all the other equations I needed to simplify but not this one. If somebody could provide a step by step solution to the equation then that would be greatly appreciated.

You want to call this an "expression". Equations have equal signs. — MathematicsStudent1122, Jun 07 '16 at 21:33

score 1 · Answer 1 · answered Jun 08 '16 at 12:26

1

Ok, so you want to simplify $$ (m^3)^4(2x^3)^7(m^2)^5(3x)^2 $$ As the other answer attempts to, you can focus on each factor on its own. You simply use the rules that

$x^ax^b = x^{a+b}$ and
$(x^a)^b = x^{ab}$ and
$(xy)^{a} = x^ay^a$

So $$ \begin{align} (m^3)^4 &= m^{3\cdot 4} = m^{12} \\ (2x^3)^7 &= 2^7(x^3)^7 = 2^7x^{21}\\ (m^2)^5 &= m^{10}\\ (3x)^2 &= 3^2x^2 \end{align} $$ So $$\begin{align} (m^3)^4(2x^3)^7(m^2)^5(3x)^2 &= m^{12}2^7x^{21}m^{10}3^2x^2 \\ &= 2^73^2m^{12}m^{10}x^{21}x^{2} \\ &= 2^73^2m^{22}x^{23} \end{align} $$

answered Jun 08 '16 at 12:26

Thomas

43,555

It's not (m^3)^2 but instead (m^3)^4. And the calculator is giving me an answer of 128x^21 for (2x^3)^7. – Anonymous Jun 08 '16 at 12:30
@Anonymous: I already fixed that :) – Thomas Jun 08 '16 at 12:30
The calculator is giving me an answer of 1152m^22x^23. – Anonymous Jun 08 '16 at 12:32
@Anonymous: Well, what is $2^7$ times $3^2$? – Thomas Jun 08 '16 at 12:32
Isn't (2x^3)^7 equal to 128^21? – Anonymous Jun 08 '16 at 12:33
@Anonymous: Nope. $(\color{red}{2}\color{green}{x^3})^7 = \color{red}{2}^7(\color{green}{x^3})^7$ because of the rule $(\color{red}{x}\color{green}{y})^n = \color{red}{x}^n\color{green}{y}^n$ – Thomas Jun 08 '16 at 12:35
I don't get it. – Anonymous Jun 08 '16 at 12:36
@Anonymous: Consider this example $6^3 = 216$. And since $6 = 2\cdot 3$ you could also do $(2\cdot3)^3 = 2^3\cdot3^3 = 8\cdot 27$ which is still $216$. – Thomas Jun 08 '16 at 12:39
Interesting. But what about (2x^3)^7=2x^32x^32x^32x^3^2x^32x^3*2x^3? – Anonymous Jun 08 '16 at 12:41
@Anonymous: Exactly, so you will have $7$ factors of $2$ and $7$ factors of $x^3$. – Thomas Jun 08 '16 at 12:49
Are you sure that's the correct answer? – Anonymous Jun 08 '16 at 13:35
@Anonymous: I am pretty sure. – Thomas Jun 08 '16 at 17:17

Simplifying $(m^3)^4(2x^3)^7(m^2)^5(3x)^2$

1 Answers1