Question

.

Answer 1

Rewrite 81 as 3 to some power

Answer 2

3^4 ???

Answer 3

3^2 =9 , 3^3 =27 and so on .

Answer 4

81^-3/4.=((3^4))-3/4=3^-3=1/27

Answer 5

The negative sign in an index/power/exponent means to invert the result

Eg 4^-1 means 1/4 or one quarter.

Look up index rules or exponents.

Answer 6

\[81^\left( -3/4 \right) = \frac{1}{81^\left(3/4 \right)}\]

Answer 7

Okay, basically, if you simplify 81^(-3/4), you'll get 1/27. Now, I know this isn't at all apparent initially, so let me step you through it.
\[81^\left( \frac{-3}{4} \right)=\frac{1}{81^\left(\frac{3}{4} \right)}\]
Both of these are equivalent, by the definition of negative powers.

Answer 8

Now, taking the square of 81 and correcting the power:
\[\frac{1}{81^\left( \frac{3}{4} \right)}= \frac{1}{9^\left( \frac{6}{4} \right)}\]
Simplifying the power fraction:
\[\frac{1}{9^\left( \frac{6}{4} \right)} = \frac{1}{9^\left( \frac{3}{2} \right)}\]
Calculating the top power out:
\[\frac{1}{9^\left( \frac{3}{2} \right)} = \frac{1}{729^\left( \frac{1}{2} \right)}\]
Taking the square root of the denominator:
\[\frac{1}{729^\left( \frac{1}{2} \right)} = \frac{1}{27}\]

Answer 9

Forgot this step:
\[\frac{1}{729^\left( \frac{1}{2} \right)} = \frac{1}{\sqrt{729}}\]
Also, in the second step, I should've just taken the square root of 81 to get:
\[\frac{1}{81^\left( \frac{3}{4} \right)} = \frac{1}{\sqrt{81}^\left( \frac{3}{2} \right)}\]
Does this make sense?