Answer

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Hint: Here we have to substitute ${{\log }_{5}}x=t$ and $x={{5}^{t}}$ and apply rule that when bases are same then exponents will be equal and then factorize the equation to get values of t and put it back to get values of x.

Complete step-by-step solution:

In the question we are given the function,

${{x}^{\left[ {{\left( {{\log }_{5}}x \right)}^{2}}-\left( 9/2 \right){{\log }_{5}}x+5 \right]}}=5\sqrt{5}\ldots \ldots (1)$

Now for solving the equation (1) we will consider ${{\log }_{5}}x=t$ then the equation (1) can be represented as

${{x}^{\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}=5\sqrt{5}\ldots \ldots (2)$

Here in the equation (2) we can replace x by ${{5}^{t}}$ as ${{\log }_{5}}x=t$ so it can be written as,

${{5}^{t\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}=5\sqrt{5}\ldots \ldots (3)$

In the right hand side of equation (3) $5\sqrt{5}$ can be represented as ${{5}^{\dfrac{3}{2}}}$ to make the bases same so that we can apply the rule of indices which is when bases are same then exponents will be equal and vice-versa hence equation (3) can be represented as,

${{5}^{t\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}={{5}^{\dfrac{3}{2}}}\ldots \ldots (4)$

Now by applying the rule of indices which we get,

${{t}^{3}}-\dfrac{9}{2}{{t}^{2}}+5t=\dfrac{3}{2}\ldots \ldots (5)$

Now we are moving $\dfrac{3}{2}$ from right hand side to left hand side of equation (5) so we get,

${{t}^{3}}-\dfrac{9}{2}{{t}^{2}}+5t-\dfrac{3}{2}=0\ldots \ldots (6)$

Now we will factorize the left hand side of equation (6) to find the value of t so it goes like,

${{t}^{3}}-{{t}^{2}}-\dfrac{7}{2}{{t}^{2}}+\dfrac{7}{2}t+\dfrac{3}{2}t-\dfrac{3}{2}=0$

which can be further written as,

${{t}^{2}}\left( t-1 \right)-\dfrac{7}{2}t\left( t-1 \right)+\dfrac{3}{2}\left( t-1 \right)=0$

Taking $\left( t-1 \right)$ common we will get,

$\left( t-1 \right)\left\{ {{t}^{2}}-\dfrac{7}{2}t+\dfrac{3}{2} \right\}=0$

Further factoring $\left( {{t}^{2}}-\dfrac{7}{2}t+\dfrac{3}{2} \right)$ we get,

$\left( t-1 \right)\left\{ {{t}^{2}}-\dfrac{1}{2}t-3t+\dfrac{3}{2} \right\}=0$

which can be further written as,

$\left( t-1 \right)\left\{ t\left( t-\dfrac{1}{2} \right)-3\left( t-\dfrac{1}{2} \right) \right\}=0$

which can finally written as,

$\left( t-1 \right)\left( t-3 \right)\left( t-\dfrac{1}{2} \right)=0$

We can say that for values t=1, 3, $\dfrac{1}{2}$ the equation (2) satisfies.

As we know the value of t we can find the value of x by using $x={{5}^{t}}$ which we took earlier for solving.

So, the value of x is ${{5}^{1}},{{5}^{3}},{{5}^{\dfrac{1}{2}}}$ which can be written as 5, 125, $\sqrt{5}$ respectively.

Now analysing the option we can say that equation A. holds as there are exactly three real solutions. For option B. it also holds as it has minimum one real solution to satisfy. Now for option C. it holds as only one irrational solution $\sqrt{5}$ satisfy the equation (1). But it does not hold for option D. as there are no complex solutions.

Hence the correct options are ‘A’, ‘B’, ‘C’.

Note: Student while taking ${{\log }_{5}}x=t$ they confuse how to replace with x so they can use ${{\log }_{a}}b=c$ can also be represented as ${{a}^{c}}=b$ hence it can be used here by ${{\log }_{5}}x=t$ as $x={{5}^{t}}$. And after finding values of t, the students forget to put it back to x as hence get mistakes while choosing options.

Complete step-by-step solution:

In the question we are given the function,

${{x}^{\left[ {{\left( {{\log }_{5}}x \right)}^{2}}-\left( 9/2 \right){{\log }_{5}}x+5 \right]}}=5\sqrt{5}\ldots \ldots (1)$

Now for solving the equation (1) we will consider ${{\log }_{5}}x=t$ then the equation (1) can be represented as

${{x}^{\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}=5\sqrt{5}\ldots \ldots (2)$

Here in the equation (2) we can replace x by ${{5}^{t}}$ as ${{\log }_{5}}x=t$ so it can be written as,

${{5}^{t\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}=5\sqrt{5}\ldots \ldots (3)$

In the right hand side of equation (3) $5\sqrt{5}$ can be represented as ${{5}^{\dfrac{3}{2}}}$ to make the bases same so that we can apply the rule of indices which is when bases are same then exponents will be equal and vice-versa hence equation (3) can be represented as,

${{5}^{t\left[ {{t}^{2}}-\dfrac{9t}{2}+5 \right]}}={{5}^{\dfrac{3}{2}}}\ldots \ldots (4)$

Now by applying the rule of indices which we get,

${{t}^{3}}-\dfrac{9}{2}{{t}^{2}}+5t=\dfrac{3}{2}\ldots \ldots (5)$

Now we are moving $\dfrac{3}{2}$ from right hand side to left hand side of equation (5) so we get,

${{t}^{3}}-\dfrac{9}{2}{{t}^{2}}+5t-\dfrac{3}{2}=0\ldots \ldots (6)$

Now we will factorize the left hand side of equation (6) to find the value of t so it goes like,

${{t}^{3}}-{{t}^{2}}-\dfrac{7}{2}{{t}^{2}}+\dfrac{7}{2}t+\dfrac{3}{2}t-\dfrac{3}{2}=0$

which can be further written as,

${{t}^{2}}\left( t-1 \right)-\dfrac{7}{2}t\left( t-1 \right)+\dfrac{3}{2}\left( t-1 \right)=0$

Taking $\left( t-1 \right)$ common we will get,

$\left( t-1 \right)\left\{ {{t}^{2}}-\dfrac{7}{2}t+\dfrac{3}{2} \right\}=0$

Further factoring $\left( {{t}^{2}}-\dfrac{7}{2}t+\dfrac{3}{2} \right)$ we get,

$\left( t-1 \right)\left\{ {{t}^{2}}-\dfrac{1}{2}t-3t+\dfrac{3}{2} \right\}=0$

which can be further written as,

$\left( t-1 \right)\left\{ t\left( t-\dfrac{1}{2} \right)-3\left( t-\dfrac{1}{2} \right) \right\}=0$

which can finally written as,

$\left( t-1 \right)\left( t-3 \right)\left( t-\dfrac{1}{2} \right)=0$

We can say that for values t=1, 3, $\dfrac{1}{2}$ the equation (2) satisfies.

As we know the value of t we can find the value of x by using $x={{5}^{t}}$ which we took earlier for solving.

So, the value of x is ${{5}^{1}},{{5}^{3}},{{5}^{\dfrac{1}{2}}}$ which can be written as 5, 125, $\sqrt{5}$ respectively.

Now analysing the option we can say that equation A. holds as there are exactly three real solutions. For option B. it also holds as it has minimum one real solution to satisfy. Now for option C. it holds as only one irrational solution $\sqrt{5}$ satisfy the equation (1). But it does not hold for option D. as there are no complex solutions.

Hence the correct options are ‘A’, ‘B’, ‘C’.

Note: Student while taking ${{\log }_{5}}x=t$ they confuse how to replace with x so they can use ${{\log }_{a}}b=c$ can also be represented as ${{a}^{c}}=b$ hence it can be used here by ${{\log }_{5}}x=t$ as $x={{5}^{t}}$. And after finding values of t, the students forget to put it back to x as hence get mistakes while choosing options.

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