Logarithm - MathsTips.com

Let ‘a’ and ‘M’ be two real numbers such that $a>0$ , $a\neq 1$ and $M>0$ . A real number $x$ is called “the logarithm of M to the base a”, denoted by $\log_a{M}$ if $a^x=M$

For example, $\log_2{8}=3$ , since $2^3=8$

A few more examples:

$\because$ $2^6=64 \therefore$ $\log_2{64}=6$
$\because$ $3^{-4}=\dfrac{1}{3^4}=\dfrac{1}{81} \therefore$ $\log_3{(\dfrac{1}{81})}=(-4)$
$\because$ $10^{-2}=\dfrac{1}{100}=0.01 \therefore$ $\log_{10}{0.01}=(-2)$

NOTE: It should be noted that the logarithm of any arbitrarily chosen real number with respect to any arbitrarily chosen base (a real number) is not defined. The essence of the definition of logarithm of a real number $M$ to the base $a$ is that it should be a unique real number. The definition of logarithm should be such that, for $M \neq1, \log_a{M}=\log_b{M}$ if and only if $a=b$ and this happens only when:

$M>0$
$a>0, a \neq1$ .

None of the above two conditions can be dropped or relaxed while defining logarithm.

Logarithm Facts

Fact 1: If $M \leq0$ and $a>0$ , then $a^x \neq M$ for any real number $x$ .

$\because$ $a^x>0$ for any real $x$ when $a>0$

Thus, the logarithm of zero, or a negative number to any positive base is undefined.

Fact 2: If $M \neq0$ and $a=0$ then, $0^x\neq M$ for any real number $x$ .

$\because$ $0^x=0$ for any real number $x$ and it is undefined for $x\leq0$

So, the logarithm of $M \neq0$ with respect to the base zero is not defined.

Fact 3: If $M=0$ and $a=0$ then there exists no real (unique) $x$ for which $a^x=M$ .

Notice that $0^2=0^3=0^4=....=0$

So, logarithm with respect to the base zero is not unique and hence logarithm of 0 to the base 0 is undefined.

Fact 4: If $M=1=a$ , we have $1^x=1$ for any real number $x$ . Thus the logarithm of 1 with respect to the base 1 is not uniquely defined.

Fact 5: If $M>0$ , but $M \neq1$ and $a=1$ , then $1^x \neq M$ for any real number $x$ . Thus, the logarithm of $M$ ( $M>0, M\neq1$ ) is not defined with respect top base 1.

Fact 6: If $M>0, a<0$ , i.e., if we allow negative numbers to be the bases of logarithms of a number $M(>0)$ , we have to accept an undesirable fact that the logarithms of a number $M(\neq1)$ , with respect to different bases can be equal.

NOTE: $\log_a{1}=\log_b{1}=0$ even when $a \neq b$ since $a^0=1$ by definition.

Let us consider the following example:

We know, $2^4=16$ and $(-2)^4=16$ as well.

So, if we allow negative numbers to be the bases of logarithms, then,

$\log_2{16}=\log_{-2}{16}=4$ ,

i.e., the logarithm of 16 with respect to two different bases 2 and (-2) are equal(=4). This brings ambiguity to the definition of logarithm.

Fact 7: If $M<0$ and $a>0$ , i.e., if the logarithm of a negative number with negative base is permitted, then,

$\because$ $(-1)^1=(-1)^{-1}$ ,

$\therefore$ $\log_{-1}{(-1)}=1$ as well as $(-1)$ ,

i.e., uniqueness of logarithm of a number with a given base would not remain valid.

For all these reasons, we consider only the positive real numbers ( $\neq1$ ) as the bases of logarithms of a real number $M(>0)$ .

NOTE: Under all the conditions for which logarithm of a number is defined, the statement the

“ $\log_aM=\log_bM$ fopr $M\neq1$ if and only if $a=b$ “,

is equivalent to the well known property of the indices that

“ $a^x=b^x$ for $x\neq0$ if and only if $a=b$ ”

Logarithm Laws

Law 1: $\log_a{MN}=\log_aM+\log_aN$

Proof: Let $\log_a{M}=x$

$\therefore$ $a^x=M$

Let $\log_a{N}=y$

$\therefore$ $a^y=N$

Now, $a^x.a^y=M.N$

or, $a^{x+y}=MN$

$\therefore$ from definition we have,

$\log_a{MN}=x+y$

$\therefore$ $\log_a{MN}=x+y=\log_a{M}+\log_a{N}$ [putting the values of x and y]

$\Rightarrow$ $\log_a{MN}=\log_a{M}+\log_a{N}$

Corollary:

This law is true for more than two positive factors.

i.e., $\log_a{MNP}=\log_a{MN} +\log_a{P}=\log_a{M} +\log_a{N} +\log_a{P}$

In general, $\log_a{MNP....}=\log_a{M} +\log_a{N} +\log_a{P} +....$

Hence, the logarithm of the product of two or more positive factors to any positive base other than 1 is equal to the sum of the logarithms of the factors to the same base.

Law 2: $\log_a{(\dfrac{M}{N})} =\log_a{M}-\log_a{N}$

Proof: Let $\log_a{M}=x$

$\therefore$ $a^x=M$

Let $\log_a{N}=y$

$\therefore$ $a^y=N$

Now, $\dfrac{a^x}{a^y}=\dfrac{M}{N}$

or, $a^{x-y}=\dfrac{M}{N}$

$\therefore$ from definition we have,

$\log_a{(\dfrac{M}{N})}=x-y$

$\therefore$ $\log_a{(\dfrac{M}{N})} =x-y =\log_a{M}-\log_a{N}$ [Putting the values of x and y]

$\Rightarrow$ $\log_a{\dfrac{M}{N})} =\log_a{M} -\log_a{N}$

Corollary:

$\log_a{(\dfrac{M \times N \times P \times ....}{R \times S \times T \times....})} \\ \vspace{5mm} \\ =\log_a{(MNP....)} -\log_a{(RST....)} \\ \vspace{5mm} \\ = (\log_a{M} +\log_a{N} +\log_a{P} +....) -(\log_a{R} +\log_a{S} +\log_a{T} +....)$

Thus, the logarithm of the quotient of two factors to any base other than 1 is equal to the difference of the logarithms of the factors to the same base.

Law 3: $\log_a{M^n}=n\log_a{M}$

Proof: Let $\log_a{M^n}=x$

$\therefore$ $a^x=M^n$

Let $\log_a{M}=y$

$\therefore$ $a^y=M$

Now, $a^x=M^n=(a^y)^n=a^{ny}$

$\therefore$ $x=ny$

or, $\log_a{M^n}=n\log_a{M}$ [Putting the values of x and y]

Thus, this law states that when finding the logarithm of a number raised to a certain power, it can be evaluated by multiplying the logarithm of the number by that power.

Law 4: $\log_a{M}=\log_b{M} \times \log_a{b}$

Proof: Let $\log_a{M}=x$

$\therefore$ $a^x=M$

Let $\log_b{M}=y$

$\therefore$ $b^y=M$

$\log_a{b}=z$ (say)

$\therefore$ $a^z=b$

Now, $a^x=M=b^y=(a^z)^y=a^{yz}$

$\therefore$ $x=yz$ % or, $\log_a{M}=\log_b{M} \times \log_a{b}$ [Putting the values of x, y, z]

Corollary:

a) Putting $M=a$ on both sides of law (4), we get,

$\log_a{a}=\log_b{a} \times \log_a{b}$

or, $\log_b{a} \times \log_a{b}=1$ [ $\because$ $\log_a{a}=1$ ]

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

i .e., the logarithm of positive number $'a'$ with respect to a positive base $'b'$ ( $\neq1$ ) is equal to the reciprocal of logarithm of $'b'$ with respect to the base $'a'$

b) From law (4) we get,

$\log_b{M}=\dfrac{\log_a{M}}{\log_a{b}}$

i.e., the logarithm of a positive number $M$ with respect to a positive base $b(\neq1)$ is equal to the quotient of the logarithm of the number $M$ and the logarithm of the number $b$ both with respect to the any positive base $a(\neq1)$ .

NOTE: If bases are not stated in the logarithms in a problem, we assume same bases for all the logarithms.

Logarithm of 1

Any number raised to the power zero is 1, $a^0=1$ . The logarithmic form of this is: $\log_a{1}=0$

The logarithm of 1 in with respect to base is zero.

Examples

Example 1: Find $\log_2{512}$

Solution: This is the same as being asked “What is 512 expressed as a power of 2?”

Now, $2^9=512$

$\therefore$ $\log_2{512}=\log_2{2^9}=9\log_2{2}=9$

Example 2: Find $\log_8{\dfrac{1}{64}}$

Solution: This is the same as being asked “What is $\dfrac{1}{64}$ expressed as a power of 8?”

Now, $\dfrac{1}{64}=64^{(-1)}=(8^2)^{(-1)}=8^{(-2)}$

$\therefore$ $\log_8{\dfrac{1}{64}}=\log_8{8^{(-2)}}=(-2)$

Example 3: If $3+\log_{10}{x}=2\log_{10}{y}$ find $x$ in terms of $y$

Solution: $3+\log_{10}{x}=2\log_{10}{y} \\ \vspace{5mm} \\ \Rightarrow 3\log_{10}{10} + \log_{10}{x}=2\log_{10}{y} \\ \vspace{5mm} \\ \Rightarrow \log_{10}{10^3} +\log_{10}{x}=\log_{10}{y^2} \\ \vspace{5mm} \\ \Rightarrow \log_{10}[10^3.x]=\log_{10}{y^2} \\ \vspace{5mm} \\ \Rightarrow 10^3.x=y^2 \\ \vspace{5mm} \\ \Rightarrow x=\dfrac{y^2}{1000}$

Example 4: Find $\log{128}-7\log{2}$

Solution: $\log{128}-7\log{2} \\ \vspace{5mm} \\ =\log{2^7}-7\log{2} \\ \vspace{5mm} \\ =7\log{2}-7\log{2} \\ \vspace{5mm} \\ =0$

Example 5: Show that $\log_2{10}-\log_8{125}=1$

$\log_8{125}=\log_8{5^3}=3\log_8{5}=3.\dfrac{1}{\log_5{8}}=3.\dfrac{1}{\log_5{2^3}}=3.\dfrac{1}{3.\log_5{2}} \\ \vspace{5mm} \\ =\dfrac{1}{\log_5{2}}=\log_2{5}$

$L.H.S. =\log_2{10}-\log_8{125} \\ \vspace{5mm} \\ =\log_2{10}-\log_2{5} \\ \vspace{5mm} \\ =\log_2{\dfrac{10}{5}} \\ \vspace{5mm} \\ =\log_2{2} \\ \vspace{5mm} \\ =1 (Proved)$

Standard Bases

There are two bases which are used much more commonly than any other base and hence deserve special mention. These are base 10 and base $e$ .

Logarithms to the base 10 or $\log_10$ re often simply written as $\log$ without explicitly writing the base. So if you see an expression like $\log{x} $ you can assume the base to be 10. Calculators are pre-designed and pre-programmed to evaluate logarithms to the base 10.

The second common base is $'e'$ . The symbol $e$ is called the exponential constant and has a value approximately equal to 2.718. This is a number like $'\pi'$ in the sense that it has an infinite decimal expansion. Base $'e'$ is used because this constant occurs frequently in the mathematical, physical, biological and economic world of applications. If you see an expression like $'lnx'$ , you can assume the base to be $'e'$ . Such logarithms are also called Naperian or Natural Logarithms.

Using logarithms to solve equations

We can use logarithms to solve equations where the unknown is in the power. Suppose we wish to solve the equation $3^x=5$ . we can solve this by taking logarithms on both sides. Whilst logarithms to any base can be used it is common practice to use base 10 as these are readily available in log tables or on calculator.

So, $\log{3^x}=\log{5} \\ \vspace{5mm} \\ \Rightarrow x\log{3}=\log {5} \\ \vspace{5mm} \\ \Rightarrow x = \dfrac{\log{5}}{\log{3}}$

This value can be calculated using a log tables or a scientific calculator.

Solve: $3^x=5^(x-2)$

Solution: The unknown appears in the power. So, we have to take logarithm on both sides.

$\log{3^x}=\log{5^{(x-2)}} \\ \vspace{5mm} \\ \Rightarrow x\log{3}=(x-2)\log{5} \\ \vspace{5mm} \\ \Rightarrow x\log{3}=x\log{5}-2\log{5} \\ \vspace{5mm} \\ \Rightarrow x\log{3}-x\log{5}=-2\log{5} \\ \vspace{5mm} \\ \Rightarrow x\log{5}-x\log{3}=2\log{5} \\ \vspace{5mm} \\ \Rightarrow x(\log{5}-\log{3})=\log{5^2} \\ \vspace{5mm} \\ \Rightarrow x\log{\dfrac{5}{3}}=\log{25} \\ \vspace{5mm} \\ \Rightarrow x=\dfrac{\log{25}}{\log{\dfrac{5}{3}}}$

Inverse Operations

Suppose we pick a base, say, 2. Suppose we pick a power, say 8.

We now raise the base to the power 8 to get $2^8$ .

Now, suppose we take logarithm to the base 2 of $2^8$ . We then have $\log_2{2^8}$ . Using the laws of logarithm this can be written as $8\log_2{2}$ .

Recall that $\log_a{a}=1$ , so $\log_2{2}=1$ and so we simply have 8, i.e., the number we had started with.

So, raising 2 to a power and taking logarithm to base 2 of the result are inverse operations.

Let us look at this in another way. Suppose we pick a number, say, 8. Suppose we dind its logarithm to the base 2, i.e., $\log_2{8}$ .

Suppose we raise the base 2 to this power, i.e., $2^{\log_2{8}}$ .

Now, $\because$ $8=2^3$ , we can write this as $2^{\log_2{2^3}}$ .

Using the laws of logarithm, $2^{\log_2{2^3}}=2^{3\log_2{2}}$ which equals $2^3=8$ since, $\log_2{2}=1$ .

We see that raising 2 to the logarithm of a number to the base 2 results in the original number. So, raising a base to a power and finding the logarithm to that base of the result are inverse operations. Doing one operation and following it up with by the other we end up where we started.

Exercise

Use logarithm to solve the following equations: a) $10^x=\dfrac{1}{2}$ b) $e^x=0.1$ c) $10^x=e^{(2x-1)}$
Show that: $\log_2{(8x)^{\dfrac{1}{3}}}=1+\dfrac{1}{3}\log_2{x}$
Find $x$ if $2\log_b{5}+\dfrac{1}{2}\log_b{9}-\log_b{3}=\log_b{x}$
Find $y$ if $\log_2{\dfrac{y}{3}}=4$
Use the laws of logarithm to solve $\log_3{x}=\log_3{7}+\log_3{3}$
If $y$ is 21, what is the value of $\log_e{e^{(2y)}}$
Simplify: $\ln({e^2\ln{3}})$

Logarithm Facts

Logarithm Laws

Logarithm of 1

Examples

Standard Bases

Inverse Operations

Exercise

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