Question

What is the value of \[\left[ {{\left( {{\left( {{x}^{y+1}} \right)}^{\dfrac{{{y}^{2}}}{{{y}^{2}}-1}}} \right)}^{1-\dfrac{1}{y}}} \right]\]?
(a). xy
(b). \[{{x}^{y}}\]
(c). \[{{y}^{x}}\]
(d). \[\dfrac{y}{x}\]

Answer 1

Hint: Use the experiment product rule to solve the expression. Substitute the basic identities and cancel out similar powers. Thus simplify to get the value of the given expression.

Complete step-by-step answer:

The given question is an algebraic problem of exponents. Thus we can solve the exponent with the help of “power rule”.
The power rule tells us that to raise a power to a power, just multiply the exponents. For example, \[{{5}^{2}}\] raised to the \[{{3}^{rd}}\] power is equal to \[{{5}^{6}}\].
The General form of power rule can be given by,
\[{{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}\]
Eg: - \[{{\left( {{5}^{2}} \right)}^{3}}={{5}^{2\times 3}}={{5}^{6}}\].
Now, consider our question, \[\left[ {{\left( {{\left( {{x}^{y+1}} \right)}^{\dfrac{{{y}^{2}}}{{{y}^{2}}-1}}} \right)}^{1-\dfrac{1}{y}}} \right]\].
This is of the form, \[{{\left( {{\left( {{x}^{m}} \right)}^{n}} \right)}^{p}}\].
Where; m = y + 1.
\[\begin{align}
  & n=\dfrac{{{y}^{2}}}{{{y}^{2}}-1} \\
 & p=1-\dfrac{1}{y} \\
\end{align}\]
We can write, \[{{\left( {{\left( {{x}^{m}} \right)}^{n}} \right)}^{p}}={{x}^{mnp}}\].
We know the basic formula, \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\].
i.e. \[{{y}^{2}}-1=\left( y-1 \right)\left( y+1 \right)\]
\[\begin{align}
  & \therefore {{x}^{mnp}}={{x}^{\left( y+1 \right)\left( \dfrac{{{y}^{2}}}{{{y}^{2}}-1} \right)\left( 1-\dfrac{1}{y} \right)}} \\
 & {{x}^{mnp}}={{x}^{\left( y+1 \right)\times \dfrac{{{y}^{2}}}{\left( y-1 \right)\left( y+1 \right)}\times \dfrac{\left( y-1 \right)}{y}}} \\
\end{align}\]
Cancel out the like terms in the power and simplify it.
Cancel out (y + 1), (y - 1) and y from the numerator and denominator.
Thus we get the find answer as \[{{x}^{y}}\].
\[\begin{align}
  & \therefore {{x}^{mnp}}={{x}^{y}} \\
 & \left[ {{\left( {{\left( {{x}^{y+1}} \right)}^{\dfrac{{{y}^{2}}}{y-1}}} \right)}^{1-\dfrac{1}{y}}} \right]={{x}^{y}} \\
\end{align}\]
\[\therefore \] Option (b) is the correct answer.

Note: This is a basic question of exponents that can be solved by power rule. Similar rules are product rule \[{{x}^{m}}.{{x}^{n}}={{x}^{mn}}\]. Quotient rule where, \[{{x}^{m}}\div {{x}^{n}}={{x}^{m-n}}\] zero rule, where \[{{x}^{0}}=1,x\ne 0\].