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 ${b^{{{\log }_b}x}} = x$ .