Question

2. Consider the following equation with radicals and rational exponents. 11^19/4 • ^a√11^b = 11^9/4 • √11^3 What is the quotient of b and a?

Answer 1

The given expression :

$11^{\frac{19}{4}}\cdot\sqrt[a]{11^b}=11^{\frac{9}{4}}\cdot\sqrt[]{11^3}$

The exponents in the form of root can be express as :

$x^{\frac{n}{m}}=\sqrt[m]{x^n}$

So, the given expression simplify as :

$\begin{gathered} 11^{\frac{19}{4}}\cdot\sqrt[a]{11^b}=11^{\frac{9}{4}}\cdot\sqrt[]{11^3} \\ 11^{\frac{19}{4}}\cdot11^{\frac{b}{a}}=11^{\frac{9}{4}}\cdot11^{\frac{3}{2}} \end{gathered}$

Since the base of the exponents are same and the bases are multiply so, the exponents will sum up

$\begin{gathered} 11^{\frac{19}{4}}\cdot11^{\frac{b}{a}}=11^{\frac{9}{4}}\cdot11^{\frac{3}{2}} \\ 11^{\frac{19}{4}+\frac{b}{a}}=11^{\frac{9}{4}+\frac{3}{2}} \end{gathered}$

Now, the bases of exponents on both side are equal so, the exponents will equate together

$\begin{gathered} 11^{\frac{19}{4}+\frac{b}{a}}=11^{\frac{9}{4}+\frac{3}{2}} \\ \frac{19}{4}+\frac{b}{a}=\frac{9}{4}+\frac{3}{2} \\ \frac{b}{a}=\frac{9}{4}+\frac{3}{2}-\frac{19}{4} \\ \frac{b}{a}=\frac{9+6-19}{4} \\ \frac{b}{a}=\frac{-4}{4} \\ \frac{b}{a}=(-1) \\ b=(-1)a \\ \text{ From the divsion alogrithm } \\ \text{Dividend = Divisor}\times Quotient+\text{ Remainder} \\ b=(-1)a \\ \text{Quotient of b =(-1)} \\ \text{Now, for a} \\ \frac{b}{a}=-1 \\ \frac{b}{a}=\frac{-1}{1} \\ \text{Apply cross multiplication} \\ b(1)=(-1)a \\ \text{Multiply by (-1)} \\ (-1)b=a \\ a=(-1)b \\ \text{From Division alogrithm} \\ \text{Quotient of a is (-1)} \end{gathered}$

Answer : Quotient of b and a is ( -1 ).