The following equation has:
${{x}^{\left[ {{\left( {{\log }_{5}}x \right)}^{2}}-\left( 9/2 \right){{\log }_{5}}x+5 \right]}}=5\sqrt{5}$
A. exactly three real solutions
B. at least one real solutions
C. exactly one irrational solution
D. complex roots
Last updated date: 20th Mar 2023
•
Total views: 305.4k
•
Views today: 2.85k
Answer
305.4k+ views
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.
Recently Updated Pages
If abc are pthqth and rth terms of a GP then left fraccb class 11 maths JEE_Main

If the pthqth and rth term of a GP are abc respectively class 11 maths JEE_Main

If abcdare any four consecutive coefficients of any class 11 maths JEE_Main

If A1A2 are the two AMs between two numbers a and b class 11 maths JEE_Main

If pthqthrth and sth terms of an AP be in GP then p class 11 maths JEE_Main

One root of the equation cos x x + frac12 0 lies in class 11 maths JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
