Question

The value of $\sum\limits_{n = 1}^m {\log \dfrac{{{a^{2n - 1}}}}{{{b^{m - 1}}}}} $($a$≠$0$, $b$≠$0$,$1$) is equal to
A. $m\log \dfrac{{{a^{2m}}}}{{{b^{m - 1}}}}$
B. $\log \dfrac{{{a^{2m}}}}{{{b^{m - 1}}}}$
C. $\dfrac{m}{2}\log \dfrac{{{a^{2m}}}}{{{b^{2m - 2}}}}$
D. $\dfrac{m}{2}\log \dfrac{{{a^{2m}}}}{{{b^{m + 1}}}}$

Answer 1

Hint: Here the summation is taking over $n = 1$ to $m$. So there are $m$ terms. To get the solution students need to use the rules and properties of logarithms to get a required solution. The value of $a$ is not equal to $0$ and $b \ne 0$. If $b = 0$ then the denominator becomes zero and we cannot proceed any more also $b \ne 1$ is given.

Formula used: Power of power rule: ${\left( {{x^m}} \right)^n} = {a^{mn}}$
Logarithm power rule law: $\log \left( {{M^n}} \right) = n\log \left( M \right)$

Complete step-by-step solution:
Let $I$=$\sum\limits_{n = 1}^m {\log \dfrac{{{a^{2n - 1}}}}{{{b^{m - 1}}}}} $
Expand the summation from $n$=$1$ to $m$
$I$=$\log \dfrac{a}{{{b^{m - 1}}}}$+$\log \dfrac{{{a^3}}}{{{b^{m - 1}}}}$+$\log \dfrac{{{a^5}}}{{{b^{m - 1}}}}$+…+\[\log \dfrac{{{a^{2m - 1}}}}{{{b^{m - 1}}}}\]$ - - - - - \left( 1 \right)$
If the terms are the same then we can add the powers. The powers of $a$ in the numerators are $1,3,5,...,2m - 1$ and in the denominator there are $m$ terms. Now adding the powers of numerator and denominator in equation $\left( 1 \right)$ becomes
$I$= \[\log \dfrac{{{a^{\left( {1 + 3 + 5 + ... + 2m - 1} \right)}}}}{{{b^{{{\left( {m - 1} \right)}^m}}}}}\]
Now, $1$+$3$+$5$+…+$2m$−$1$=${m^2}$
For, ${S_m}$=$\dfrac{m}{2}$$\left( {a + l} \right)$
Where $a$ is the first term and $l$ is the last term.
${S_m} = \dfrac{m}{2}\left( {1 + 2m - 1} \right)$
Simplifying we get,
$ \Rightarrow {S_m} = \dfrac{m}{2}\left( {2m} \right)$
Cancelling the similar terms in numerator and denominator,
$ \Rightarrow {S_m} = {m^2}$
Hence,
$\therefore $ $1 + 3 + 5 + ... + 2m - 1 = {m^2}$
We get, $I$= $\log \dfrac{{{a^{{m^2}}}}}{{{b^{\left( {m - 1} \right)}}^m}}$
$ \Rightarrow I$=$\log \dfrac{{{a^{2m}}}}{{{b^{{{\left( {m - 1} \right)}^m}}}}} - - - - - \left( 2 \right)$
In equation $\left( 2 \right)$, we used the power of power rule in the numerator.
Power of power rule:
For any number $x$ and all integers $m$ and $n$; ${\left( {{x^m}} \right)^n} = {a^{mn}}$
From equation $\left( 2 \right)$, $I = \log \left( {\dfrac{{{{\left( {{a^{2m}}} \right)}^{\dfrac{m}{2}}}}}{{{{\left( {{b^{\left( {m - 1} \right)2}}} \right)}^{\dfrac{m}{2}}}}}} \right) - - - - - \left( 3 \right)$
In equation $\left( 3 \right)$, just multiply and divide by $2$ in the powers of numerator and denominator.
Now, $I = \log {\left( {\dfrac{{{a^{2m}}}}{{\left( {{b^{2m - 2}}} \right)}}} \right)^{\dfrac{m}{2}}} - - - - - \left( 4 \right)$
$ \Rightarrow $$I = \dfrac{m}{2}\log \dfrac{{{a^{2m}}}}{{{b^{2m - 2}}}} - - - - - \left( 5 \right)$
In equation $\left( 5 \right)$ we used power rule of logarithm that is, $\log \left( {{M^n}} \right) = n\log \left( M \right)$
$\therefore $ $I = \dfrac{m}{2}\log \left( {\dfrac{{{a^{2m}}}}{{\left( {{b^{2m - 2}}} \right)}}} \right)$
$\therefore {\text{ }}\sum\limits_{n = 1}^m {\log \dfrac{{{a^{{2^{n - 1}}}}}}{{{b^{m - 1}}}}} {\text{ = }}\dfrac{m}{2}\log \dfrac{{{a^{2m}}}}{{{b^{2m - 2}}}}$

The answer is option $\left( C \right)$

Note: The laws which we used in the problems are power of power rule, law of power in logarithm. Also there are some more logarithm laws: product rule law, quotient rule law, power rule law and change of base rule law. Also the important formula we used in the problem is $1 + 3 + 5 + ..... + 2m - 1 = {m^2}$.