# Math Help - Evaluating this fraction

1. ## Evaluating this fraction

How would I evaluate the following fraction and express it properly?

${\frac{2^6}{\displaystyle(\frac{1}{2^{3x}})}}$

Should I multiply and take the reciprocal of the bottom or would the brackets block me from taking the reciprocal?

2. Originally Posted by Kataangel

Should I multiply and take the reciprocal of the bottom
Yep

Originally Posted by Kataangel

would the brackets block me from taking the reciprocal?
Nah

3. Disregard brackets...there just for show

4. $
2^6\times 2^{3x} = 2^{6+3x}
$

5. Don't get picky, pickslides

6. Originally Posted by Wilmer
Disregard brackets...there just for show
Actually, they're not just there for show, without them the expression is ambiguous.

7. Yeah that's why I thought that you couldn't reciprocate them.

8. Well, what you can do is write it like this:

$\frac {2^6}{1} \div \frac 1 {2^{3x}}$

which is what the initial expression means.

Then by the rule "to divide by a fraction you turn it upside down and multiply instead" you then get:

$\frac {2^6}{1} \times \frac {2^{3x}} 1$

Alternatively you can take the original expression:

$\frac {2^6}{\left({\dfrac 1 {2^{3x}}}\right)}$

... and multiply top and bottom by the reciprocal of what's on the bottom:

$\frac {2^6 \times 2^{3x}}{\left({\dfrac 1 {2^{3x}}}\right) \times 2^{3x}}$

... which is probably the clearest way to explain what's going on.

9. Originally Posted by Kataangel
How would I evaluate the following fraction and express it properly?

${\frac{2^6}{\displaystyle(\frac{1}{2^{3x}})}}$

Should I multiply and take the reciprocal of the bottom or would the brackets block me from taking the reciprocal?
You can also say that $\frac{1}{2^{3x}} = 2^{-3x}$ so the expression reduces to

$\frac{2^6}{2^{-3x}}$

and then apply the laws of exponents:

$\frac{2^6}{2^{-3x}} = 2^{6-(-3x)} = 2^{6+3x}$

To go further you can factorise the exponent:

$2^{6+3x} = 2^{3(2+x)}$

And applying the laws again

$2^{3(2+x)} = (2^3)^{2+x} = 8^{2+x}$