Operations With Logs

We got introduced to logarithm . In this topic we will see some basic operations of log and understand their derivations.

$\underline{I.\ Log\ Of\ Products}$

If we have a product of two numbers $x$ and $y$, then we the following relationship holds:

$\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$

We will see the derivation here:

Let: $\log_b{(x)} = p$ and $\log_b{(y)} = q$ $... (i)$

$\therefore b^p = x$ and $b^q = y$

multiplying both these equations, we get:

$b^p \times b^q = xy$

$\Rightarrow b^{p + q} = xy$

$\Rightarrow \log_b{(xy)} = p + q$ now we replace the values of $p$ and $q$ from step $(i)$, and get:

$\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$

Expanding this rule for $n$ numbers, we get:

$\log_b{(a_1 \cdot a_2 \cdot a_3 ... a_n)} = \log_b{(a_1)} + \log_b{(a_2)} + \log_b{(a_3)} + ... + \log_b{(a_n)}$

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$\underline{II.\ Log\ Of\ Ratios}$

If we have a ratio of two numbers, $\dfrac{x}{y}$, then the following relationship holds:

$\log_b{\left(\dfrac{x}{y}\right)} = \log_b{(x)} - log_b{(y)}$

We will use a similar strategy as above, to derive this relationship.

Let: $\log_b{(x)} = p$ and $\log_b{(y)} = q$ $... (i)$

$\therefore b^p = x$ and $b^q = y$

Dividing both the equations, we get:

$\dfrac{b^p}{b^q} = \dfrac{x}{y}$

$\Rightarrow b^{p - q} = \dfrac{x}{y}$

$\Rightarrow \log_b{\left(\dfrac{x}{y}\right)} = p - q$

Replacing the values of $p$ and $q$ from $(i)$ we get:

$\log_b{\left(\dfrac{x}{y}\right)} = \log_b{(x)} - \log_b{(y)}$

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$\underline{III.\ Log\ Of\ Powers}$

$\log_{b}{(x^y)} = y\log_{b}{(x)}$

We will derive the above relationship in the following way:

Let $\log_{b}{(x)} = p$

$\therefore b^p = x$

raising both sides to the power $y$ we get:

$\left(b^p\right)^y = x^y$

$\Rightarrow b^{yp} = x^y$

$\Rightarrow log_{b}{(x^y)} = yp$

replacing $p = log_{b}{(x)}$ in the above equation, we get:

$\log_{b}{(x^y)} = y\log_{b}{(x)}$

$\underline{IV.\ Reciprocal\ Of\ Logs}$

If $a$ and $b$ are positive real numbers, and $a \ne 1$ and $b \ne 1$, then:

$\log_b{a} = \dfrac{1}{\log_a{b}}$

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We will derive this using the following steps:

Let $\log_b{a} = p$

$\therefore b^p = a$

$\Rightarrow b = a^{\dfrac{1}{p}}$

$\Rightarrow \log_a{b} = \dfrac{1}{p}$

$\Rightarrow p = \dfrac{1}{\log_a{b}}$

Now, replacing the value $p = \log_b{a}$, we get:

$\log_b{a} = \dfrac{1}{\log_a{b}}$

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$\underline{V.\ Chain\ Rule\ Of\ Logs}$

For real numbers $a$, $b$ and $c$, we have:

$\log_b{(a)} \times \log_c{(b)} = \log_c{(a)}$

To derive this, we will use the following method:

Let $\log_b{(a)} = p$ and $\log_c{(b)} = q$

$\therefore b^p = a$ $...(i)$ and

$c^q = b$ $...(ii)$

From $(i)$ we get $b = a^{\frac{1}{p}}$ and from $(ii)$ we have $b = c^q$

$\therefore c^q = a^{\frac{1}{p}}$

$\Rightarrow c^{pq} = a$

$\Rightarrow log_c{a} = pq$

Now substituting the original values of $p$ and $q$ we get:

$\log_c{a} = \log_b{(a)} \times \log_c{(b)}$

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$\underline{VI.\ Change\ Of\ Base\ Rule}$

For real numbers $a$, $b$ and $c$, we have:

$\log_b{a} = \dfrac{\log_c{a}}{\log_c{b}}$

We will derive this identity using the rules we have specified earlier:

Using the chain rule from $(V)$, we get:

$\log_b{a} = \log_c{a} \times \log_b{c}$

Now, using the reciprocal rule from $(IV)$ we can write $\log_b{c} = \dfrac{1}{\log_c{b}}$, which gives us:

$\log_b{a} = \log_c{a} \times \dfrac{1}{\log_c{b}}$

$\Rightarrow \log_b{a} = \dfrac{\log_c{a}}{\log_c{b}}$