Title: Default for 6.3 Logarithms
Course: HELM Workbooks 2023-2024

Logarithms

Logarithms reverse the process of raising a base '\(a\)' to a power '\(n\)'.

As with all exponentials, the base should be a positive number. If \(b=a^n\) then we write \(\log_ab=n\).

Of course, the reverse statement is equivalent: If \(\log_ab=n\), then \(b=a^n\).

The expression \(log_ab=n\) is read: “The log to base \(a\) of the number \(b\) is equal to \(n\)".

The term “log" is short for the word logarithm.

Example

Determine the log equivalents of

1. \(16 = 2^4\)
2. \(16 = 4^2\)
3. \(1000 = 10^3\)
4. \(134.896=10^{2.13}\)
5. \(8.414867=e^{2.13}\)

Solution

1. Since \(16=2^4\), \(\log_216=4\).
2. Since \(16=4^2\), \(\log_416=2\).
3. Since \(1000=10^3\), \(\log_{10}1000=3\).
4. Since \(134.896=10^{2.13}\), \(\log_{10}134.896=2.13\).
5. Since \(8.414867=e^{2.13}\), \(\log_e8.414867=2.13\).

Summary:

Key Point

If \(b=a^n\), then \(log_ab=n\).

If \(\log_ab=n\), then \(b=a^n\).

Exercise

Calculate the following.

1. \(\log_{[value]}[value] = \)
2. \(\displaystyle{\log_{[value]}[value] =} \)
Feedback: Since \( \log_a a^n = n\) by definition, we have \(\log_{[value]}[value] = \log_{[value]}[value]^[value] = [value]\), \(\log_{[value]}[value] = \log_{[value]}[value]^[value] = [value]\).

Laws of logarithms

From the previous exercise (see feedback), we see that the index law \(a^na^m=a^{n+m}\) has an equivalent logarithm law \(\log_a(bc)=\log_ab+\log_ac\). In words: "The log of a product is the sum of logs."

Indeed this property is one of the major advantages of using logarithms.They transform a product of numbers (a relatively difficult operation) to a sum of numbers (a relatively easy operation).

Each index law has an equivalent logarithm law, true for any base, listed in the following Key Point:

Key Point

The laws of logarithms | The laws of indices
\(\log_a(AB) = \log_aA + \log_aB\) | \(a^{A}a^B = a^{A+B} \)
\(\log_a \left( \dfrac{A}{B} \right) = \log_aA - \log_aB \) | \(\dfrac{a^A}{a^B} = a^{A-B}\)
\(\log_a(A^k) = k\log_aA\) | \((a^A)^k = a^{kA}\)
\(\log_a(a^A) = A\) | \(a^{\log_aA} = A\)
\(\log_aa = 1\) | \(a^1 = a\)
\(\log_a1 = 0\) | \(a^0 = 1\)

Simplifying expressions involving logarithms

To simplify an expression involving logarithms their laws, given in the previous Key Point, need to be used.

Example

Simplify: \(\log_{10}2-\log_{10}4+\log_{10}(4^2)+\log_{10}(\frac{10}{4})\)

Solution

The third term \(\log_{10}(4^2)\) simplifies to \(2\log_{10}4\) and the last term \(\log_{10}\left(\dfrac{10}{4}\right)=\log_{10}10-\log_{10}4=1-\log_{10}4\). So the given expression equals

\[\log_{10}2-\log_{10}4+2\log_{10}4+1-\log_{10}4 = \log_{10}2+1.\]

Logs to base 10 and natural logs

In practice only two kinds of logarithms are commonly used, those to base 10, written \(\log_{10}\) and those to base \(e\), written \(\log_{e}\) or more usually \(\ln\) or \(\log\) (called natural logarithms).

Most scientific calculators will determine the logarithm to base \(10\) and to base \(e\).

For example, \(\log_{10} 13=1.11394\) (implying \(10^{1.11394}=13\)), \(\log 23 =\ln 23=3.13549\) (implying \(e^{3.13549}=23\)).

Changing base in logarithms

It is sometimes required to express the logarithm with respect to one base in terms of a logarithm with respect to another base.

Now \(b=a^n\) implies \(\log_ab=n\) where we have used logs to base \(a\). If for some reason we want to use another base, \(p\) say, then we do as follows.

We take logs to base \(p\) of both sides of \(b=a^n\), which gives \(\log_p(b)= \log_p(a^n)=n\log_pa\) (using one of the logarithm laws), so \(n=\dfrac{\log_p(b)}{\log_p(a)}\). But we also have \(\log_ab=n\).

Therefore, we conclude \(\log_ab=\dfrac{\log_p(b)}{\log_p(a)}.\)

This is the rule to be used when converting logarithms from one base to another.

Key Point

\[\log_ab=\dfrac{\log_p(b)}{\log_p(a)}.\]

For base 10 logs: \[\log_ab=\dfrac{\log (b)}{\log (a)}.\]

For natural logs: \[\log_ab=\dfrac{\ln (b)}{\ln (a)}.\]

Example

For example, \(\log_37\) cannot be determined directly on your calculator since logs to base 3 are not available but it can be found using the above formulas (changing to base \(10\) or base \(e\)). \[\log_37=\dfrac{\log 7}{\log 3}=\dfrac{0.8450980}{0.4771212}=1.7712437\] (Check, on your calculator, that \(3^{1.7712437}=7\)).

Alternatively, \[\log_37=\dfrac{\ln 7}{\ln 3}=\dfrac{1.9459101}{1.0986123}=1.7712437\]

Example

Simplify the expression \(10^{\log_{10} x}\).

Solution

Let \(y=10^{\log_{10} x}\), then take logs (to base \(10\)) of both sides: \[\log_{10} y =\log_{10} (10^{\log_{10} x}) =(\log_{10} x)\log_{10} 10,\] where we have used: \(\log_{10} A^k=k\log_{10} A\). However, since we are using logs to base \(10\), \(\log_{10} 10=1\) and so \(\log_{10} y=\log_{10} x\), which implies \(y=x\).

Therefore, finally we conclude that \(10^{\log_{10} x}=x\).

This is an important result, true for logarithms of any base. It follows from the basic definition of the logarithm.

Key Point

Raising to the power and taking logs are inverse operations:\[a^{\log_a x}=x,\\ \log_a(a^x) =x.\]

Exercise

Simplify the following expression.

Syntax hint: Enter lg for logs base 10, e.g. if the answer is \(\log_{10}2\), then enter lg(2).

\(\displaystyle{[value] - [value] + [value] =}\)
Feedback: We have: \([value] = \log_{10}10^{[value]} = [value]\), \([value] = \log_{10}(\frac{10}{[value]}) = \log_{10}10 - \log_{10}[value] = 1 - \log_{10}[value]\), \([value] = \log_{10}10^{[value]} = [value]\). Therefore, the expression is equal to \( [value] - (1 - \log_{10}[value]) + [value] = [value]\).

Exercise

Evaluate the following.

1. \([value] =\)
2. \([value] =\)
3. \([value] =\)
Feedback: Since \(\log_aA^k = k \log_a A\), we have \([value] = \log_{10} 10^[value] = [value]\), \([value] = \log_{10} 10^[value] = [value]\), \([value] = \displaystyle [value]\).

Exercise

Evaluate the following (use a calculator if necessary).

1. \([value] =\) (Round up and enter \(4\) decimal places. If the answer has less than \(4\) decimal places, then enter zero(s) at the end.)
2. \([value] =\) (Enter a fraction.)
3. \([value] =\) (Round up and enter \(4\) decimal places. If the answer has less than \(4\) decimal places, then enter zero(s) at the end.)
Feedback: Use a calculator for (a) and (c). We get: \([value] = [value] \text{ and } [value] = [value].\) For (b), using \( \log_e e^k = k \), we get \([value] = \displaystyle [value].\)

Exercise

Re-express \(\log_{[value]} [value]\) using base \(10\) and base \(e\).

Syntax hint: Enter lg for logs base \(10\), and ln for natural logs (base \(e\)), e.g. enter lg(2) for \(\log_{10}2\), enter ln(2) for \(\log_e2\).

1. Base \(10\)
\(\log_{[value]} [value] = \)
2. Base \(e\)
\( \log_{[value]} [value] = \)
3. Now, use your calculator to determine the value of \(\log_{[value]} [value]\).
\(\log_{[value]} [value] = \) (Round up and enter \(4\) decimal places. If the answer has less than \(4\) decimal places, then enter zero(s) at the end.)
Feedback: Recall the formula for converting logarithms from one base to another \[ \log_a b = \dfrac{\log_p b}{\log_p a}. \] (a) Applying the above formula for \(a = [value], b =[value] \) and \(p = 10\), we get \[ \log_{[value]} [value] = \dfrac{\log_{10} [value]}{\log_{10}[value]}.\] (b) Applying the same formula for \(a = [value], b=[value]\) and \(p = e\), we get \[ \log_{[value]} [value] = \dfrac{\ln [value]}{\ln[value]}.\] (c) Note that \(\log_{[value]}[value]\) cannot be determined directly on your calculator since logs to base \([value]\) are not available, but it can be found by changing to base \(10\) or base \(e\). Using logs base \(10\) on your calculator, we get \[ \log_{[value]} [value] = \dfrac{\log_{10} [value]}{\log_{10}[value]} = [value].\] Alternatively, using logs base \(e\) on your calculator, we get \[ \log_{[value]} [value] = \dfrac{\ln [value]}{\ln[value]} = [value].\]

Exercise

Let \( \log_ab = [value]\) and \( \log_ac = [value]\). Express \( \log_a(bc) \) in terms of \([value]\) and \([value]\).

\( \log_a(bc) \) =
Feedback: Since \([value] = \log_ab\), \(b = a^{[value]}\). Similarly, \([value] = \log_ac\) implies \(c = a^{[value]}\). Hence, \(bc = a^{[value]}a^{[value]} = a^{[value]+[value]}\), which implies \(\log_a(bc) = [value]+[value]\). Therefore, using the property of indices, we have shown that \[\log_a(bc)=[value]+[value]=\log_ab+\log_ac.\]

Exercise

Find the values of the following.

1. \( \log_{[value]} [value] = \)
2. \( [value]^{\log_{[value]}[value]} = \)
Feedback: 1. \(\log_{[value]}[value] = \log_{[value]} [value]^{[value]} = [value].\) 2. Since raising to the power and taking logs are inverse operations (previous key point), we have: \( \displaystyle [value]^{\log_{[value]}[value]} = [value] .\)

Exercise

Evaluate the following. Round up and enter your answers in 4 decimal places. If the answer has less than 4 decimal places, then enter zero(s) at the end.

1. \( \log_{[value]} [value] = \)
2. \( \ln [value] = \)
Feedback: (a) Note that \(\log_{[value]}[value]\) cannot be determined directly on your calculator since logs to base \([value]\) are not available, but it can be found by changing to base \(10\) or base \(e\). Using logs base \(10\) on your calculator, we get \[ \log_{[value]} [value] = \dfrac{\log_{10} [value]}{\log_{10}[value]} = [value].\] Alternatively, using logs base \(e\) on your calculator, we get \[ \log_{[value]} [value] = \dfrac{\ln [value]}{\ln[value]} = [value].\] (b) Using log base \(e\) on your calculator, we get \(\ln [value] = [value]\).

Exercise

Simplify (just simplify - don't compute using a calculator) the following.

\( [value] \log_{10}[value] - [value] \log_{10}[value] + [value] \log_{10}[value]=\)

Syntax hint: Enter lg for logs base 10, e.g. enter lg(2) for \(\log_{10}2\).
Feedback: \( [value] \log_{10}[value] - [value] \log_{10}[value] + [value] \log_{10}[value]\) \(= [value] \log_{10} [value]^[value] - [value] \log_{10} [value]^[value] + [value] \log_{10} [value]^[value]\) \(= [value] \cdot [value] \log_{10} [value] - [value] \cdot [value] \log_{10}[value] + [value] \cdot [value] \log_{10}[value]\) \( = [value].\)

Exercise

Simplify (just simplify - don't compute using a calculator) the following.

Syntax hint: Enter ln for natural logs (base \(e\)), e.g. enter ln(2) for \(\log_e2\).

\([value] = \)
Feedback: \(\begin{align} [value] & = [value] \ln (10^[value]) \cdot \frac{\ln [value]}{\ln 10} - [value] \ln [value] \\ &= [value] \cdot [value] \ln 10 \cdot \frac{\ln [value]}{\ln 10} - [value] \ln [value] \\ & = [value]\end{align}. \)

Exercise

Simplify the following. Syntax hint: Enter lg(x) for \(\log_{10}x\).

\( [value] = \)
Feedback: \([value] = [value] \cdot [value] \log_{10}[value] - [value] \cdot [value] \log_{10}[value] = [value]\)

Exercise

Simplify the following. Syntax hint: Enter ln(x) or log(x) for \(\log_{e}x\).

\( [value] = \)
Feedback: \( [value] = \ln \left[ \dfrac{[value] \cdot ([value])}{[value]} \right] = [value] \)