Question

Let \[x = {4^{{{\log }_2}\sqrt {{9^{k - 1}} + 7} }}\] and \[y = \dfrac{1}{{{{32}^{\log {}_2\sqrt[5] {{{3^{k - 1}} + 1}}}}}}\] and xy=4, then the sum of the cubes of the real values (s) of k is
A.1
B.5
C.8
D.9

Answer 1

Hint: In this question the value of x and y are given and the relation between the x and y are also given so by using the logarithm power rule and the inverse property we will further solve the equation and then find the value of k.

Complete step-by-step answer:
Given functions,
\[x = {4^{{{\log }_2}\sqrt {{9^{k - 1}} + 7} }}\]
\[y = \dfrac{1}{{{{32}^{\log {}_2\sqrt[5] {{{3^{k - 1}} + 1}}}}}}\]
\[xy = 4 - - (i)\]
We can write (i) as by substituting x and y
\[
 xy = 4 \\
\Rightarrow \dfrac{{{4^{{{\log }_2}\sqrt {{9^{k - 1}} + 7} }}}}{{{{32}^{\log {}_2\sqrt[5] {{{3^{k - 1}} + 1}}}}}} = 4 \\
 \]
Now as we know \[4 = {2^2}\] and \[32 = {2^5}\] so we can further write the equation as
\[\dfrac{{{2^{2{{\log }_2}{{\left( {{9^{k - 1}} + 7} \right)}^{\dfrac{1}{2}}}}}}}{{{2^{5\log {}_2{{\left( {{3^{k - 1}} + 1} \right)}^{\dfrac{1}{5}}}}}}} = 4\]
Now by using the logarithm power rule (\[{\log _a}{x^p} = p{\log _a}x\] ) we can further write the above equation as
\[
\Rightarrow \dfrac{{{2^{2 \times \dfrac{1}{2}{{\log }_2}\left( {{9^{k - 1}} + 7} \right)}}}}{{{2^{5 \times \dfrac{1}{5}\log {}_2\left( {{3^{k - 1}} + 1} \right)}}}} = 4 \\
\Rightarrow \dfrac{{{2^{{{\log }_2}\left( {{9^{k - 1}} + 7} \right)}}}}{{{2^{\log {}_2\left( {{3^{k - 1}} + 1} \right)}}}} = 4 \;
 \]
Now by applying the inverse property of logarithm (\[{b^{{{\log }_b}x}} = x\] ) in the above obtained equation we can further write
\[
\Rightarrow \dfrac{{{2^{{{\log }_2}\left( {{9^{k - 1}} + 7} \right)}}}}{{{2^{\log {}_2\left( {{3^{k - 1}} + 1} \right)}}}} = 4 \\
\Rightarrow \dfrac{{{9^{k - 1}} + 7}}{{{3^{k - 1}} + 1}} = 4 \;
 \]
Hence by further solving (cross multiplying) this we get
\[
\Rightarrow {9^{k - 1}} + 7 = 4\left( {{3^{k - 1}} + 1} \right) \\
  {\left( 3 \right)^{2\left( {k - 1} \right)}} + 7 = 4\left( {{3^{k - 1}} + 1} \right) \\
\Rightarrow {3^{2k}} \cdot {3^{ - 2}} + 7 = 4\left( {{3^k} \cdot {3^{ - 1}} + 1} \right) \\
  {\left( {{3^k}} \right)^2} \cdot {3^{ - 2}} + 7 = 4\left( {{3^k} \cdot {3^{ - 1}} + 1} \right) \\
 \]
Now let\[{3^k} = p\] , so we can further write the equation as
\[{p^2} \cdot {3^{ - 2}} + 7 = 4\left( {p \cdot {3^{ - 1}} + 1} \right)\]
Hence by further solving, we get
\[
\Rightarrow \dfrac{{{p^2}}}{9} + 7 = 4\left( {\dfrac{p}{3} + 1} \right) \\
\Rightarrow \dfrac{{{p^2}}}{9} + 7 = \dfrac{4}{3}p + 4 \\
\Rightarrow \dfrac{{{p^2}}}{9} - \dfrac{4}{3}p + 3 = 0 \\
  {p^2} - 12p + 27 = 0 \;
 \]
Now we solve the obtained quadratic equation to find the value of p,
\[
\Rightarrow {p^2} - 12p + 27 = 0 \\
\Rightarrow {p^2} - 3p - 9p + 27 = 0 \\
\Rightarrow p\left( {p - 3} \right) - 9\left( {p - 3} \right) = 0 \\
\Rightarrow \left( {p - 3} \right)\left( {p - 9} \right) = 0 \;
 \]
Hence we get the values of \[p = 3,9\]
Now, since \[{3^k} = p\] , hence we get the value of k as
When \[p = 3\]
So,\[{3^k} = {3^1}\]
Therefore \[k = 1\]
When \[p = 9\]
So, \[{3^k} = {3^2}\]
Therefore \[k = 2\]
Hence the option which satisfies the value of k is Option A.
So, the correct answer is “Option A”.

Note: Students often confuse the power rule and the inverse rule of the logarithmic function. Power rule of the logarithmic function is \[{\log _a}{x^p} = p{\log _a}x\] while the inverse rule of the logarithmic function is \[{b^{{{\log }_b}x}} = x\] .