Question

How do you solve $27={{x}^{\dfrac{3}{2}}}$ ?

Answer 1

Hint: To solve the question, we need to know that ${{x}^{\dfrac{a}{b}}}$ is equal bth root of x to the power a. ${{\left( {{x}^{\dfrac{a}{b}}} \right)}^{\dfrac{b}{a}}}$ is equal to x. we know that ${{x}^{\dfrac{3}{2}}}$ is equal to 27. We can take the power of $\dfrac{2}{3}$ both sides to find the value of x.

Complete step by step solution:
The given equation is $27={{x}^{\dfrac{3}{2}}}$ , we know that ${{\left( {{x}^{\dfrac{a}{b}}} \right)}^{\dfrac{b}{a}}}$ is equal to x , so the value of ${{\left( {{x}^{\dfrac{3}{2}}} \right)}^{\dfrac{2}{3}}}$ is equal to 1.
By taking power of $\dfrac{2}{3}$ both sides we get ${{\left( 27 \right)}^{\dfrac{2}{3}}}=x$
We know that ${{x}^{\dfrac{a}{b}}}$ is equal bth root of x to the power a , so ${{\left( 27 \right)}^{\dfrac{2}{3}}}$ is equal to square of cube root of 27. Cube root of 27 is 3 and the square of 3 is equal to 9. So x is equal to 9.
We can check whether our answer is correct or not by putting x equal to 9 in the equation ${{x}^{\dfrac{3}{2}}}$
${{\left( 9 \right)}^{\dfrac{3}{2}}}$ is equal to a cube of square root of 9 which is a cube of 3 that is 27.
So ${{\left( 9 \right)}^{\dfrac{3}{2}}}$ is equal to 27 and 9 is the correct answer.

Note: In exponential function ${{a}^{x}}$ if the value of a is a negative real number then graph of ${{a}^{x}}$ is not continues because any fraction power of a may not exist when a is negative. So the value of a should be positive to make the graph continuous. If f(a,x ) is equal to ${{a}^{x}}$ then f( a ,- x) is equal to $f\left( \dfrac{1}{a},x \right)$ where a is not equal to 0.