Question

If ${\log _2}x + {\log _4}x + {\log _8}x = {\log _k}x$ , for all $x \in {R^ + }$ and $k = \sqrt[b]{a}$ , where $a,b \in N$ . Find ${\left( {a + b} \right)_{\min }}$ .

Answer 1

Hint : Use the formulae logarithmic base change rule i.e. ${\log _b}x = \dfrac{{{{\log }_c}x}}{{{{\log }_c}b}}$ , the logarithmic form of ${a^m} = n$ is ${\log _a}n = m$ . In addition to this, the formula i.e. ${a^{m \cdot n}} = {\left( {{a^m}} \right)^n}$ can also be used to solve this question.

Complete step-by-step answer :
${\log _2}x + {\log _4}x + {\log _8}x = {\log _k}x$ is the given equation.
We know that according to the logarithm base change rule, ${\log _b}x = \dfrac{{{{\log }_c}x}}{{{{\log }_c}b}}$
Compare the term ${\log _2}x$ with ${\log _b}x$ we get, $x = x,{\rm{ }}b = 2$ , also consider $c = 10$ .
Substitute the value of $x = x,{\rm{ }}b = 2$ and $c = 10$ in the equation ${\log _b}x = \dfrac{{{{\log }_c}x}}{{{{\log }_c}b}}$ .
$\Rightarrow {\log _2}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}}$
We know that when the argument is 10, it is not shown in the symbol. This means, ${\log _{10}}m = \log {\rm{ }}m$ .
So, ${\log _2}x = \dfrac{{{{\log }_{10}}x}}{{{{\log }_{10}}2}}$ can be written as ${\log _2}x = \dfrac{{\log x}}{{\log 2}}$
Similarly,
${\log _4}x = \dfrac{{\log x}}{{\log 4}}$ and ${\log _8}x = \dfrac{{\log x}}{{\log 8}}$
We know that $\log {a^m} = m \cdot \log a$ as per the logarithm rule.
We can write $\log 4$ as:
$\log 4 = \log {2^2}\\\ = 2\log 2......\left( 1 \right)$
Similarly, $\log 8$ can be written as:
$\log 8 = \log {2^3}\\\ = 3\log 2......\left( 2 \right)$

$${\log _2}x + {\log _4}x + {\log _8}x = {\log _k}x\\\ \dfrac{{\log x}}{{\log 2}} + \dfrac{{\log x}}{{\log 4}} + \dfrac{{\log x}}{{\log 8}} = {\log _k}x$$

After substituting the value of $\log 8 = 3\log 2$ and $\log 4 = 2\log 2$ from equations (1) and (2) in the equation $\dfrac{{\log x}}{{\log 2}} + \dfrac{{\log x}}{{\log 4}} + \dfrac{{\log x}}{{\log 8}} = {\log _k}x$, we get
$\Rightarrow \dfrac{{\log x}}{{\log 2}} + \dfrac{{\log x}}{{2\log 2}} + \dfrac{{\log x}}{{3\log 2}} = {\log _k}x$
After taking $\dfrac{{\log x}}{{\log 2}}$ common from the left-hand side of the equation, we get
$\Rightarrow \dfrac{{\log x}}{{\log 2}}\left( {1 + \dfrac{1}{2} + \dfrac{1}{3}} \right) = {\log _k}x$
Now, we have to simplify the equation by taking the denominator as LCM of 1, 2 and 3 and accordingly adjusting other values.

$$\Rightarrow \dfrac{{\log x}}{{\log 2}}\left( {\dfrac{{6 + 3 + 2}}{6}} \right) = {\log _k}x\\\ \Rightarrow \dfrac{{\log x}}{{\log 2}} \cdot \dfrac{{11}}{6} = {\log _k}x$$

On dividing the numerator and denominator of the left-hand side of the equation by 11, we get

$$\Rightarrow \left( {\dfrac{{{\raise0.7ex\hbox{ $ {11} $ } \\!{\left/ {\vphantom {{11} {11}}}\right.} \\!\lower0.7ex\hbox{ $ {11} $ }}}}{{{\raise0.7ex\hbox{ $ 6 $ } \\!{\left/ {\vphantom {6 {11}}}\right.} \\!\lower0.7ex\hbox{ $ {11} $ }}}}} \right) \cdot \dfrac{{\log x}}{{\log 2}} = {\log _k}x\\\ \dfrac{{\log x}}{{\dfrac{6}{{11}}\log 2}} = {\log _k}x$$

We know that $\log {a^m} = m \cdot \log a$ .
After comparing $\dfrac{6}{{11}}\log 2$ with $m \cdot \log a$ , we get, $m = \dfrac{6}{{11}},{\rm{ }}a{\rm{ = 2}}$ .
Substitute $m = \dfrac{6}{{11}},{\rm{ }}a{\rm{ = 2}}$ in the equation $m \cdot \log a = \log {a^m}$ .
$\Rightarrow \dfrac{6}{{11}} \cdot \log 2 = \log {2^{\left( {\dfrac{6}{{11}}} \right)}}......\left( 3 \right)$
After substituting equation (3) in the equation $\dfrac{{\log x}}{{\dfrac{6}{{11}}\log 2}} = {\log _k}x$, we get
$\dfrac{{\log x}}{{\log {2^{\left( {\dfrac{6}{{11}}} \right)}}}} = {\log _k}x$
We know that $\dfrac{{{{\log }_c}x}}{{{{\log }_c}b}} = {\log _b}x$ .
Now we have to compare $\dfrac{{\log x}}{{\log {2^{\left( {\dfrac{6}{{11}}} \right)}}}}$ with $\dfrac{{{{\log }_c}x}}{{{{\log }_c}b}}$ to find the values of $c,{\rm{ }}b,{\rm{ }}x$ .
On comparing, we get, $c = 10$ , $x = x$ and $b = {2^{\left( {\dfrac{6}{{11}}} \right)}}$ .
Substitute the values of $c = 10$ , $x = x$ and $b = {2^{\left( {\dfrac{6}{{11}}} \right)}}$ in the equation $\dfrac{{{{\log }_c}x}}{{{{\log }_c}b}} = {\log _b}x$ .
$\Rightarrow \dfrac{{\log x}}{{\log {2^{\left( {\dfrac{6}{{11}}} \right)}}}} = {\log _{{2^{\left( {\dfrac{6}{{11}}} \right)}}}}x......\left( 4 \right)$
Now, we can use the equation (4) in the equation $\dfrac{{\log x}}{{\log {2^{\left( {\dfrac{6}{{11}}} \right)}}}} = {\log _k}x$.
${\log _{{2^{\left( {\dfrac{6}{{11}}} \right)}}}}x = {\log _k}x$
In order to find the value of $k$ , we have to compare the left- and right-hand side of the equation.
$k = {2^{\left( {\dfrac{6}{{11}}} \right)}}$
As per the question, we know that $k = \sqrt[b]{a}$
Now, we can equate $k = {2^{\left( {\dfrac{6}{{11}}} \right)}}$ and $k = \sqrt[b]{a}$ to each other.
${2^{\left( {\dfrac{6}{{11}}} \right)}} = \sqrt[b]{a}$
We know that $\sqrt[b]{a}$ can also be written as ${a^{\dfrac{1}{b}}}$ .
${2^{\left( {\dfrac{6}{{11}}} \right)}} = {a^{\dfrac{1}{b}}}$
We know that ${2^{m \cdot n}} = {\left( {{2^m}} \right)^n}$ .
In order to simplify the calculation, we have to apply the formula ${2^{m \cdot n}} = {\left( {{2^m}} \right)^n}$ in the equation ${2^{\left( {\dfrac{6}{{11}}} \right)}} = {a^{\dfrac{1}{b}}}$ .
$\Rightarrow {\left( {{2^6}} \right)^{\dfrac{1}{{11}}}} = {a^{\dfrac{1}{b}}}$
${2^6} = 2 \times 2 \times 2 \times 2 \times 2 \times 2\\\ = 64$
We have to substitute equation (5) in the equation${\left( {{2^6}} \right)^{\dfrac{1}{{11}}}} = {a^{\dfrac{1}{b}}}$ to reach to the final solution.

$$\Rightarrow {\left( {{2^6}} \right)^{\dfrac{1}{{11}}}} = {a^{\dfrac{1}{b}}}\\\ \Rightarrow {\left( {64} \right)^{\dfrac{1}{{11}}}} = {a^{\dfrac{1}{b}}}$$

To find the values of $a$ and $b$ , we have to compare both sides of the equation ${\left( {64} \right)^{\dfrac{1}{{11}}}} = {a^{\dfrac{1}{b}}}$.
$\Rightarrow a = 64$ and $b = 11$
$\Rightarrow {\left( {a + b} \right)_{{\rm{minimum}}}} = 64 + 11\\\ = 75$
Therefore, the required value is ${\left( {a + b} \right)_{{\rm{minimum}}}} = 75$ .

Note : In this type of questions, students often forget to apply the logarithmic rule for base switch and base change rule. While using ${a^{m \cdot n}} = {\left( {{a^m}} \right)^n}$ formula student needs to take care that ${a^{m \cdot n}} = {\left( {{a^m}} \right)^n}$ formula is valid and ${a^{m \cdot n}} \ne {a^m} \cdot {a^n}$ .