Evaluating Logarithms

Concept

Logarithm

A logarithm is the inverse of an exponential function. The logarithm of a positive number $m$ can be written as follows. The expression $lo g_{b} (m)$ is read as $lo g$ base $b$ of $m$ and states that raising $b$ to the $n^{th}$ power yields $m .$

$lo g_{b} (m) = n \Leftrightarrow b^{n} = m$

Here, $b$ is both the base of the logarithm and the base of the exponent. For example, the value of $n$ in $lo g_{4} (16) = n$ is given by the exponent to which $4$ would be raised to result in $16 .$ $lo g_{4} (16) = n lo g_{4} (16) = 2 \Leftrightarrow 4^{n} = 16 \Leftrightarrow 4^{2} = 16$ The logarithmic form and exponential form are equivalent.

The relationship between base and exponent for logarithms and powers

Concept

Common Logarithm

A common logarithm is a logarithm of base $10 .$ For example, $lo g_{10} (1000)$ is equal to $3$ because $1 0^{3}$ is equal to $1000 .$

Since $lo g_{10}$ is used so often, it is sometimes written without a base. For positive values of $m,$ the common $lo g$ of $m$ can be defined as follows.

$lo g (m) = n \Leftrightarrow 1 0^{n} = m$

Concept

The Natural Logarithm

A natural logarithm is a logarithm with base $e$ . This means that $ln (m)$ equals the exponent to which $e$ must be raised to equal $m .$

$ln (m) = n \Leftrightarrow m = e^{n}$

Although it is correct to write $lo g_{e},$ the natural logarithm is more commonly written as $ln .$

ln (m) = lo g_{e} (m)

Rewrite the logarithmic and exponential expressions

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Exercise

For the following expressions, rewrite them in either logarithmic or exponential form. $lo g (100) = 2 8^{2} = 64 lo g_{2} (32) = 5 e^{0} = 1$

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Solution

We will rewrite each expression one at a time using the relationship $lo g_{b} (m) = n \Leftrightarrow b^{n} = m .$ Notice that the first and third are written as logarithms and the second and fourth are written as exponents. Since a base is not written on the first logarithm, we know it's base $10 .$ Furthermore, $2$ is the exponent. Thus, $lo g (100) = 2 \Leftrightarrow 1 0^{2} = 100 .$ The same reasoning applies for the third expression, $lo g_{2} (32) = 5 .$ $lo g_{2} (32) = 5 \Leftrightarrow 2^{5} = 32$ Using the same relationship in the opposite way, we can rewrite the second and fourth expression. In $8^{2} = 64,$ it can be seen that $8$ is the base and $2$ is the exponent. This means that $64$ is the value of which we take the logarithm. $8^{2} = 64 \Leftrightarrow lo g_{8} (64) = 2 .$ For the last expression, this yields $e^{0} = 1 \Leftrightarrow lo g_{e} (1) = 0 \Leftrightarrow ln (1) = 0 .$ To summarize, the following expressions are equivalent. $lo g (100) = 2 8^{2} = 64 lo g_{2} (32) = 5 e^{0} = 1 \Leftrightarrow 1 0^{2} = 100 \Leftrightarrow lo g_{8} (64) = 2 \Leftrightarrow 2^{5} = 32 \Leftrightarrow ln (1) = 0$

Evaluate the logarithms

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Exercise

Evaluate the following logarithms. $lo g_{3} (81) ln (e) lo g (\frac{1}{1 0 0 0})$

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Solution

When evaluating logarithms, it can be helpful to think about what the expression means. $lo g_{3} (81)$ asks the exponent to which $3$ must be raised to equal $81 .$ Since, $3 \cdot 3 \cdot 3 \cdot 3 = 81,$ $lo g_{3} (81) = 4 .$ The second expression, $ln (e),$ is a logarithm of base $e .$ It asks the exponent to which $e$ must be raised to equal $e .$ Since $e^{1} = e,$ $ln (e) = 1 .$ Notice that, unlike the other expressions, the last, which is of base $10,$ contains a fraction. Thus, we must consider the exponent to which $10$ is raised to equal $\frac{1}{1 0 0 0} .$ As it turns out, negative exponents yield fractions. $x^{- n} = \frac{1}{x ^{n}} .$ It can be helpful to rewrite $1000$ as a power of $10 .$ Since $1000 = 1 0^{3},$ $\frac{1}{1 0 0 0} = \frac{1}{1 0 ^{3}} = 1 0^{- 3} .$ Thus, $lo g (\frac{1}{1 0 0 0}) = - 3 .$

Evaluating Logarithms

Concept

Logarithm

Concept

Common Logarithm

Concept

The Natural Logarithm

Example

Rewrite the logarithmic and exponential expressions

Example

Evaluate the logarithms