
If $$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} },\ A^{BC}+B^{AC}+C^{AB}=729$$. Find the value of $$A^{\dfrac{1}{A} }$$.
Answer
588k+ views
Hint: In this question it is given that $$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} },\ A^{BC}+B^{AC}+C^{AB}=729$$. We have to find the value of $$A^{\dfrac{1}{A} }$$. So to find the solution we first need to express A in terms of B and similarly express C in terms of B and we have to put in the second equation from where we will get the value of B and after that we can easily find the value of $$A^{\dfrac{1}{A} }$$.
Complete step-by-step solution:
Given,
$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$..................(1)
Taking first and second from the above equation, we get,
$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$
$$\Rightarrow \left( A^{\dfrac{1}{A} }\right)^{AB} =\left( B^{\dfrac{1}{B} }\right)^{AB} $$ [ taking power AB on the both side of the equation]
$$\Rightarrow \left( A\right)^{\dfrac{1}{A} \times AB} =\left( B\right)^{\dfrac{1}{B} \times AB} $$ [since, $$\left( a^{n}\right)^{m} =a^{n\times m}$$]
$$\Rightarrow A^{B}=B^{A}$$...........(2)
Now taking second and third form the equation(1) we get,
$$B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$
$$\Rightarrow \left( B^{\dfrac{1}{B} }\right)^{BC} =\left( C^{\dfrac{1}{C} }\right)^{BC} $$ [ taking Power BC on the both side]
$$\Rightarrow \left( B\right)^{\dfrac{1}{B} \times BC} =\left( C\right)^{\dfrac{1}{C} \times BC} $$
$$\Rightarrow B^{C}=C^{B}$$
$$\Rightarrow C^{B}=B^{C}$$...............(3)
Now another equation that is given,
$$A^{BC}+B^{AC}+C^{AB}=729$$
$$\Rightarrow \left( A^{B}\right)^{C} +B^{AC}+\left( C^{B}\right)^{A} =729$$.............(4)
Now putting the value of $$A^{B}\ \text{and} \ C^{B}$$ in the above equation we get,
$$ \left( B^{A}\right)^{C} +B^{AC}+\left( B^{C}\right)^{A} =729$$
$$\Rightarrow B^{AC}+B^{AC}+B^{AC}=729$$
$$\Rightarrow 3B^{AC}=729$$
$$\Rightarrow B^{AC}=\dfrac{729}{3}$$
$$\Rightarrow B^{AC}=243$$
$$\Rightarrow B^{AC}=3^{5}$$
So by comparing the sides we can say that the above equation is satisfied if and only if,
B=3 and AC=5
Now putting the value of B in equation (1) we get,
$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$
$$\Rightarrow A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$ [taking the first two]
$$\Rightarrow A^{\dfrac{1}{A} }=3^{\dfrac{{}1}{3} }$$ [since B=3]
Which is our required solution.
Note: You can also solve this problem by taking $$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }=k$$, so from here you can able to find the value of A, B and C and after putting these values in the second equation you will get the value of k. Which helps you to find the value of $$A^{\dfrac{1}{A} }$$.
Complete step-by-step solution:
Given,
$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$..................(1)
Taking first and second from the above equation, we get,
$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$
$$\Rightarrow \left( A^{\dfrac{1}{A} }\right)^{AB} =\left( B^{\dfrac{1}{B} }\right)^{AB} $$ [ taking power AB on the both side of the equation]
$$\Rightarrow \left( A\right)^{\dfrac{1}{A} \times AB} =\left( B\right)^{\dfrac{1}{B} \times AB} $$ [since, $$\left( a^{n}\right)^{m} =a^{n\times m}$$]
$$\Rightarrow A^{B}=B^{A}$$...........(2)
Now taking second and third form the equation(1) we get,
$$B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$
$$\Rightarrow \left( B^{\dfrac{1}{B} }\right)^{BC} =\left( C^{\dfrac{1}{C} }\right)^{BC} $$ [ taking Power BC on the both side]
$$\Rightarrow \left( B\right)^{\dfrac{1}{B} \times BC} =\left( C\right)^{\dfrac{1}{C} \times BC} $$
$$\Rightarrow B^{C}=C^{B}$$
$$\Rightarrow C^{B}=B^{C}$$...............(3)
Now another equation that is given,
$$A^{BC}+B^{AC}+C^{AB}=729$$
$$\Rightarrow \left( A^{B}\right)^{C} +B^{AC}+\left( C^{B}\right)^{A} =729$$.............(4)
Now putting the value of $$A^{B}\ \text{and} \ C^{B}$$ in the above equation we get,
$$ \left( B^{A}\right)^{C} +B^{AC}+\left( B^{C}\right)^{A} =729$$
$$\Rightarrow B^{AC}+B^{AC}+B^{AC}=729$$
$$\Rightarrow 3B^{AC}=729$$
$$\Rightarrow B^{AC}=\dfrac{729}{3}$$
$$\Rightarrow B^{AC}=243$$
$$\Rightarrow B^{AC}=3^{5}$$
So by comparing the sides we can say that the above equation is satisfied if and only if,
B=3 and AC=5
Now putting the value of B in equation (1) we get,
$$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }$$
$$\Rightarrow A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }$$ [taking the first two]
$$\Rightarrow A^{\dfrac{1}{A} }=3^{\dfrac{{}1}{3} }$$ [since B=3]
Which is our required solution.
Note: You can also solve this problem by taking $$A^{\dfrac{1}{A} }=B^{\dfrac{1}{B} }=C^{\dfrac{1}{C} }=k$$, so from here you can able to find the value of A, B and C and after putting these values in the second equation you will get the value of k. Which helps you to find the value of $$A^{\dfrac{1}{A} }$$.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Trending doubts
Which places in India experience sunrise first and class 9 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write the 6 fundamental rights of India and explain in detail

Difference Between Plant Cell and Animal Cell

What is the Full Form of ISI and RAW

What is pollution? How many types of pollution? Define it


