Question

If the third term in the expansion of ${\left[ {x + {x^{{{\log }_{10}}x}}} \right]^5}$ is ${10^6}$, then x can be
(This question has multiple choice options)
A. ${10^{ - 1/3}}$
B. $10$
C. ${10^{ - 5/2}}$
D. ${10^2}$

Answer 1

We use the concept of binomial expansion and count the value of r which gives the third term of the expansion. Using base 10, solve the logarithm values using the formulas of log. Form a quadratic equation in the end and use factorization method to solve for the value.
A binomial expansion helps us to expand expressions of the form ${(a + b)^n}$ through the formula ${(a + b)^n} = \sum\limits_{r = 0}^n {^n{C_r}{{(a)}^r}{{(b)}^{n - r}}}$
Formula of combination is given by $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, where factorial is expanded by the formula $n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1$
$\log (mn) = \log m + \log n$ and $\log ({m^n}) = n\log m$

Complete step-by-step solution:
We are given the term ${\left[ {x + {x^{{{\log }_{10}}x}}} \right]^5}$
Use binomial expansion to expand this bracket, here $a = x,b = {x^{{{\log }_{10}}x}},n = 5$
${\left[ {x + {x^{{{\log }_{10}}x}}} \right]^5}{ = ^5}{C_0}{(x)^0}{({x^{{{\log }_{10}}x}})^{5 - 0}}{ + ^5}{C_1}{(x)^1}{({x^{{{\log }_{10}}x}})^{5 - 1}}{ + ^5}{C_2}{(x)^2}{({x^{{{\log }_{10}}x}})^{5 - 2}} + ..{ + ^5}{C_5}{(x)^5}{({x^{{{\log }_{10}}x}})^{5 - 5}}$
Third term of the expansion is $^5{C_2}{(x)^2}{({x^{{{\log }_{10}}x}})^{5 - 2}}$..............… (1)
Since we are given the value of third term is ${10^6}$
Equate the value of third term from equation (1)
${ \Rightarrow ^5}{C_2}{(x)^2}{({x^{{{\log }_{10}}x}})^{5 - 2}} = {10^6}$
Use the formula of combination to solve LHS
$\Rightarrow \dfrac{{5!}}{{3!2!}}{(x)^2}{({x^{{{\log }_{10}}x}})^3} = {10^6}$
Write numerator using factorial expansion
$\Rightarrow \dfrac{{5 \times 4 \times 3!}}{{3!2!}}{(x)^2}{({x^{{{\log }_{10}}x}})^3} = {10^6}$
Cancel same terms from numerator and denominator
$\Rightarrow 10{(x)^2}{({x^{{{\log }_{10}}x}})^3} = {10^6}$
Divide both sides by 10
$\Rightarrow \dfrac{{10{{(x)}^2}{{({x^{{{\log}_{10}}x}})}^3}}}{{10}} = \dfrac{{{{10}^6}}}{{10}}$
Cancel same terms from numerator and denominator
$\Rightarrow {(x)^2}{({x^{{{\log }_{10}}x}})^3} = {10^5}$...............… (2)
We apply ${\log _{10}}$on both sides of the equation
\Rightarrow {\log _{10}}\left\\{ {{{(x)}^2}{{({x^{{{\log }_{10}}x}})}^3}} \right\\} = {\log _{10}}\left\\{ {{{10}^5}} \right\\}
Since we law of log tells us $\log (mn) = \log (m) + \log (n)$
\Rightarrow {\log _{10}}\left\\{ {{{(x)}^2}} \right\\} + {\log _{10}}\left\\{ {{{({x^{{{\log }_{10}}x}})}^3}} \right\\} = {\log _{10}}\left\\{ {{{10}^5}} \right\\}
Also, law of log gives $\log ({m^n}) = n\log (m)$
\Rightarrow 2{\log _{10}}(x) + 3{\log _{10}}({x^{{{\log }_{10}}x}}) = 5{\log _{10}}\left\\{ {10} \right\\}
We know log base 10 cancels 10
$\Rightarrow 2{\log _{10}}(x) + 3{\log _{10}}({x^{{{\log }_{10}}x}}) = 5$
Also, law of log gives $\log ({m^n}) = n\log (m)$
\Rightarrow 2{\log _{10}}(x) + 3\left\\{ {{{\log }_{10}}x({{\log }_{10}}x)} \right\\} = 5
$\Rightarrow 2{\log _{10}}(x) + 3{({\log _{10}}x)^2} - 5 = 0$.................… (3)
Put $({\log _{10}}x) = y$
Equation (3) converts into
$\Rightarrow 3{y^2} + 2y - 5 = 0$
We can write $2y = - 3y + 5y$
$\Rightarrow 3{y^2} - 3y + 5y - 5 = 0$
Take common factors
$\Rightarrow 3y(y - 1) + 5(y - 1) = 0$
$\Rightarrow (3y + 5)(y - 1) = 0$
Equate both factors to zero
$\Rightarrow 3y + 5 = 0$
Shift constant to RHS
$\Rightarrow 3y = - 5$
Divide both sides by 3
$\Rightarrow y = \dfrac{{ - 5}}{3}$.....................… (4)
Also, $y - 1 = 0$
Shift constant to RHS
$\Rightarrow y = 1$..................… (5)
We know $({\log _{10}}x) = y$
$\Rightarrow ({\log _{10}}x) = 1$
We can write $1 = {\log _{10}}\left( {10} \right)$
$\Rightarrow {\log _{10}}\left( x \right) = {\log _{10}}\left( {10} \right)$
Cancel log base 10 on both sides of the equation
$\Rightarrow x = 10$

$\therefore$Option B is the correct answer.

Note: Students are likely to get confused while applying log rules as they are not familiar with base 10. Students are advised to compare the laws of general logarithm with base 10 logarithms as they are similar. Many students make mistakes when shifting values from one side to another, always changing the sign from positive to negative and vice versa when shifting values from one side to other side.