Question

Simplify the following.
a. $ - 27 = {\left( { - 3} \right)^x}$
b. $\dfrac{{81}}{{121}} = {\left( {\dfrac{9}{{11}}} \right)^y}$
c. $\dfrac{9}{{36}} = {\left( {\dfrac{x}{6}} \right)^2}$
d. ${x^4} = \dfrac{1}{{81}}$

Answer 1

Hint: First find the squares and cubes of the given actual fractions and integers without variables. When they are written in squares, fourth powers and cubes, if the bases are equal on either side then the powers must be equal. Equate the powers and find the values of variables.

Complete step-by-step answer:
We are given four equations, one side it has a fraction or an integer and the other side its exponential form with unknown exponents and numerators. We have to find the unknown values.
a. $ - 27 = {\left( { - 3} \right)^x}$
LHS has -27 and RHS has -3 raised to the power x. So we have to find whether -27 can be written in exponential form where the base is -3 or not.
-27 can be written as 9 times -3 and 9 can be written -3 times -3.
Which results,
$ - 27 = 9 \times - 3 = - 3 \times - 3 \times - 3 = {\left( { - 3} \right)^3}$
As we can see the bases are equal in LHS and RHS of ${\left( { - 3} \right)^3} = {\left( { - 3} \right)^x}$
So the powers must be equated, which results $3 = x
\Rightarrow x = 3$
The value of x is 3.
b. $\dfrac{{81}}{{121}} = {\left( {\dfrac{9}{{11}}} \right)^y}$
LHS has $\dfrac{{81}}{{121}}$ and RHS has $\dfrac{9}{{11}}$ raised to the power y. So we have to find whether $\dfrac{{81}}{{121}}$ can be written in exponential form where the base is $\dfrac{9}{{11}}$ or not.
81 can be written as 9 times 9 and 121 can be written 11 times 11.
Which results,
$\dfrac{{81}}{{121}} = \dfrac{{9 \times 9}}{{11 \times 11}} = \dfrac{{{9^2}}}{{{{11}^2}}} = {\left( {\dfrac{9}{{11}}} \right)^2}$
As we can see the bases are equal in LHS and RHS of ${\left( {\dfrac{9}{{11}}} \right)^2} = {\left( {\dfrac{9}{{11}}} \right)^y}$
So the powers must be equated, which results $2 = y
\Rightarrow y = 2$
 c. $\dfrac{9}{{36}} = {\left( {\dfrac{x}{6}} \right)^2}$
Here 9 can be written as 3 times 3 and 36 can be written as 6 times 6.
$\dfrac{9}{{36}} = \dfrac{{3 \times 3}}{{6 \times 6}} = \dfrac{{{3^2}}}{{{6^2}}} = {\left( {\dfrac{3}{6}} \right)^2}$
On comparing LHS and RHS, we get
$
  {\left( {\dfrac{3}{6}} \right)^2} = {\left( {\dfrac{x}{6}} \right)^2} \\
  \therefore x = 3 \\
 $
d. ${x^4} = \dfrac{1}{{81}}$
1 can be written as base 1 raised to the power 4 and 81 can be written as base 3 raised to the power 4.
$\dfrac{1}{{81}} = \dfrac{{{1^4}}}{{{3^4}}} = {\left( {\dfrac{1}{3}} \right)^4}$
On comparing LHS and RHS, we get
$
  {x^4} = {\left( {\dfrac{1}{3}} \right)^4} \\
  \therefore x = \dfrac{1}{3} \;
 $

Note: A positive exponent means repeated multiplication of the base with the base, a negative exponent means repeated division of the base by the base. When the negative exponents are moved to the denominator, the exponent becomes positive and vice-versa. Be careful with the signs of the exponents and bases as 27 is not the same as -27 and 3 is not the same as -3.