How would I evaluate the following fraction and express it properly?
$\displaystyle {\frac{2^6}{\displaystyle(\frac{1}{2^{3x}})}}$
Should I multiply and take the reciprocal of the bottom or would the brackets block me from taking the reciprocal?
How would I evaluate the following fraction and express it properly?
$\displaystyle {\frac{2^6}{\displaystyle(\frac{1}{2^{3x}})}}$
Should I multiply and take the reciprocal of the bottom or would the brackets block me from taking the reciprocal?
Well, what you can do is write it like this:
$\displaystyle \frac {2^6}{1} \div \frac 1 {2^{3x}}$
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:
$\displaystyle \frac {2^6}{1} \times \frac {2^{3x}} 1$
hence the answer.
Alternatively you can take the original expression:
$\displaystyle \frac {2^6}{\left({\dfrac 1 {2^{3x}}}\right)}$
... and multiply top and bottom by the reciprocal of what's on the bottom:
$\displaystyle \frac {2^6 \times 2^{3x}}{\left({\dfrac 1 {2^{3x}}}\right) \times 2^{3x}}$
... which is probably the clearest way to explain what's going on.
You can also say that $\displaystyle \frac{1}{2^{3x}} = 2^{-3x}$ so the expression reduces to
$\displaystyle \frac{2^6}{2^{-3x}}$
and then apply the laws of exponents:
$\displaystyle \frac{2^6}{2^{-3x}} = 2^{6-(-3x)} = 2^{6+3x}$
To go further you can factorise the exponent:
$\displaystyle 2^{6+3x} = 2^{3(2+x)}$
And applying the laws again
$\displaystyle 2^{3(2+x)} = (2^3)^{2+x} = 8^{2+x}$