Question

Multiply x¾ X ⅛

A. x^1/8
B. x^3/10
C. x^11/12
D. x^2/5

Anyone? i'm not sure if i'm supposed to multiply the fractions or cross multiply..?

Answer 1

anyonee?

Answer 2

Powers to powers, multiply. when multiplying a set of powers you add.

Answer 3

That just completely confused me. So, i'm supposed to add them?
i don't get any of the answers given though.

Answer 4

you add them, think of taking 5^(1/2) * 5^(1/2) aka the square root times the square root, you know that two square roots multiplied gives you back just 5. So then think do you add 1/2+1/2 to get 5^1 or multiply. because 5^(1/4) doesn't equal 5. then you know it's based on adding

Answer 5

the answer should be x^(7/8) idk why it's not in your choices

Answer 6

Well those are the only answers i am given, so is it possible to do this another way?

Answer 7

unfortunately no. it's a math rule. you can try it with other numbers and prove it's adding. 14^(1/4)* 14^(4/11) = 5.050168167 also written 14^(27/44)

Answer 8

OH. wait a second, instead of the ^1/8 it's ^1/6

Answer 9

then it's x^11/12

Answer 10

So the answer is x^1/8 correct?

Answer 11

oh, okay. you sure?

Answer 12

yes it's x^(11/12)

Answer 13

Okay, thank youu. can you help me with one more? i'll post it in another question

Answer 14

Look. These are all the Exponent Laws you need to know:
\[A^x \times A^y=A^{x+y}\]\[\frac{ A^x }{ A^y }=A^{x-y}|A^y \ne 0\]\[(A^x)^y=A^{xy}\]\[A^xB^x=(AB)^x\]\[\frac{ A^x }{ B^x }=\left( \frac{ A }{ B } \right)^x\]\[A^0=1|A \ne 0\]\[A^{-x}=\frac{ 1 }{ A^x }|A^x \ne 0\]You're question is x^(3/4) * x^(1/8). To solve this, we use the Product Exponent Law, which is the first one in this list. Using it we get:\[A^x \times A^y=A^{x+y} \implies x^{3/4} \times x^{1/8} =x^{(3/4)+(1/8)}=x^{7/8}\]Get it?

@brazukinhaa

Answer 15

es, and i got 4/10 but that's not an option..
And instead of 1/8 it's 1/6