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: Here, we will first rewrite the powers of given equation and then we will use the power rule, \[{\left( {{a^b}} \right)^c} = {a^{bc}}\] in the obtained equation. Then we will simplify the equation to find the required value.

Complete step by step solution: We are given that the equation is \[\left[ {{{\left( {{{\left( {{x^{y + 1}}} \right)}^{\dfrac{{{y^2}}}{{{y^2} - 1}}}}} \right)}^{1 - \dfrac{1}{y}}}} \right]\].
Rewriting the powers of the above equation, we get
\[ \Rightarrow \left[ {{{\left( {{{\left( {{x^{y + 1}}} \right)}^{\dfrac{{{y^2}}}{{{y^2} - 1}}}}} \right)}^{\dfrac{{y - 1}}{y}}}} \right]\]
Using the power rule, \[{\left( {{a^b}} \right)^c} = {a^{bc}}\] in the above equation, we get
\[ \Rightarrow \left[ {{{\left( {{x^{y + 1}}} \right)}^{\dfrac{{{y^2}}}{{{y^2} - 1}} \times \dfrac{{y - 1}}{y}}}} \right]\]
Using the rule, \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] in the above equation, we get
\[
   \Rightarrow \left[ {{{\left( {{x^{y + 1}}} \right)}^{\dfrac{y}{{\left( {y - 1} \right)\left( {y + 1} \right)}} \times \dfrac{{y - 1}}{1}}}} \right] \\
   \Rightarrow \left[ {{{\left( {{x^{y + 1}}} \right)}^{\dfrac{y}{{\left( {y + 1} \right)}}}}} \right] \\
 \]
Using the power rule, \[{\left( {{a^b}} \right)^c} = {a^{bc}}\] in the above equation, we get
\[
   \Rightarrow \left[ {{x^{\left( {y + 1} \right) \times \dfrac{y}{{\left( {y + 1} \right)}}}}} \right] \\
   \Rightarrow {x^y} \\
 \]

Hence, option B is correct.

Note:
We need to use the basic properties like \[{\left( {{a^b}} \right)^c} = {a^{bc}}\] and \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] while solving the problem and be thorough with them, or else the final answer can be wrong. Since the equation is not given to be equal to 0, so the logarithm rule won’t apply here. The function \[{x^a}\] means \[x\] is multiplied \[a\] times. When the value is same in the numerator and denominator then it cancels each other.