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
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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} $ .
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} $ .
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