Answer
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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 $$A^{\dfrac{1}{A} }$$.
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