Logarithms as Inverse Exponentials
Logarithm
For $a>0$, $a\neq 1$, and $x>0$, $\log_a x$ is the exponent $k$ such that $a^k=x$.
- This definition immediately gives the key restrictions:
- The base must satisfy $a>0$ and $a\neq 1$.
- The argument (the input to the log) must satisfy $x>0$.
- Two bases are especially common:
- Common logarithm: $\log x$ means $\log_{10} x$.
- Natural logarithm: $\ln x$ means $\log_e x$, where $e\approx 2.71828$.
- Historically, logarithms were developed to simplify computations involving very large numbers (John Napier, 1614).
- Modern calculators compute logs directly, but the algebraic rules remain central for solving problems.
Converting between Logarithmic and Exponential Form
- If you can switch forms fluently, most log questions become much easier.
- $\log_a x = y$ means $a^y = x$.
- $a^y = x$ means $\log_a x = y$.
- Rewrite $\log_2 8 = 3$ in exponential form.
- Since "log equals exponent", $2^3=8$.
Logarithm Laws and Why They Work
The log laws mirror the exponent laws because logarithms are defined through exponentials.
Product Rule
Product rule
$\log_a(xy)=\log_a x+\log_a y$.
Proof sketch:
- Let $p=\log_a x$ and $q=\log_a y$. Then $a^p=x$ and $a^q=y$.
- Multiply: $$a^p\cdot a^q=a^{p+q}=xy$$
- Taking $\log_a$ of both sides gives $\log_a(xy)=p+q=\log_a x+\log_a y$.
Quotient Rule
Quotient rule
$\log_a\left(\frac{x}{y}\right)=\log_a x-\log_a y$.
Proof sketch:
- Using the same definitions, $$\frac{x}{y}=\frac{a^p}{a^q}=a^{p-q}$$
- So $$\log_a\left(\frac{x}{y}\right)=p-q=\log_a x-\log_a y$$
Power Rule
Power rule
$\log_a(x^n)=n\log_a x$ (for real $n$, as long as $x^n$ is defined and $x>0$).
Proof sketch:
- If $p=\log_a x$ then $x=a^p$. Raise both sides to power $n$: $$x^n=(a^p)^n=a^{np}$$
- So $$\log_a(x^n)=np=n\log_a x$$
The power rule is particularly important because it lets you "bring an exponent down" in front of a logarithm.
- A very common mistake is to split logs across addition or subtraction: $$\log_a(M+N)\neq \log_a M+\log_a N$$
- Only products, quotients, and powers simplify nicely.
Expanding and Condensing Logarithmic Expressions
Many questions ask you to either:
- Expand a logarithm into a sum/difference of simpler logs (often linear logs), or
- Condense a sum/difference of logs into a single logarithm.
Expanding to a Sum and Difference of Linear Logs
A typical strategy is:
- Factor the expression inside the log as much as possible.
- Use the product and quotient rules to split factors.
- Use the power rule to bring exponents down.
Expand $\ln(x^2-9)$ into a sum/difference of linear logarithms.
Solution
- Factor first: $$x^2-9=(x-3)(x+3)$$
- Then: $$\ln(x^2-9)=\ln(x-3)+\ln(x+3)$$ with the understanding that the domain must make each factor positive.
Expand $\ln\left(\frac{x+1}{x^2-4}\right)$ into linear logarithms.
Solution
- Factor the denominator: $$x^2-4=(x-2)(x+2)$$
- Then: $$\ln\left(\frac{x+1}{x^2-4}\right)=\ln(x+1)-\ln(x-2)-\ln(x+2)$$
Writing Expressions in Terms of Given Logs
Sometimes you are given substitutions like $a=\ln x$, $b=\ln(x-1)$, $c=\ln 3$, and you must rewrite more complex expressions using $a,b,c$.
Given $a=\ln x$ and $b=\ln(x-1)$, express $\ln\left(\frac{x}{x-1}\right)$ in terms of $a$ and $b$.
Solution
Use the quotient rule: $$\ln\left(\frac{x}{x-1}\right)=\ln x-\ln(x-1)=a-b$$
- When you see a square root, rewrite it as a power first: $\sqrt{A}=A^{1/2}$.
- Then the power rule applies immediately.
Condensing Multiple Logs into One
To combine logs into a single log:
- Use the power rule to move coefficients into exponents.
- Turn sums into products (product rule reversed).
- Turn differences into quotients (quotient rule reversed).
Express $3\ln x + 2\ln(2x+1)$ as a single logarithm.
Solution
- Power rule: $$3\ln x=\ln(x^3), \qquad 2\ln(2x+1)=\ln\big((2x+1)^2\big)$$
- Product rule: $$3\ln x + 2\ln(2x+1)=\ln\left(x^3(2x+1)^2\right)$$
- If you expand $(2x+1)^2$ to $4x^2+4x+1$, do not forget the factor $x^3$ outside.
- Many algebra errors happen after the log step, not during it.
Change of Base and Using Log Tables
Change of base formula
For $a>0$, $a\neq 1$, $b>0$, and any convenient base $c>0$, $c\neq 1$:
$$\log_a b=\frac{\log_c b}{\log_c a}$$
Common choices are $c=10$ (using $\log$) or $c=e$ (using $\ln$).
- This explains how people used base-10 log tables to compute logs in other bases.
- If you can look up $\log_{10} 5$ and $\log_{10} 2$, you can compute: $$\log_2 5 = \dfrac{\log 5}{\log 2}$$
Use table values $\log 5\approx 0.69897$ and $\log 2\approx 0.30103$ to estimate $\log_2 5$.
Solution
$$\log_2 5 \approx \dfrac{0.69897}{0.30103}\approx 2.32$$
This agrees with a calculator (to 3 s.f.).
- In non-calculator settings, the change of base formula is often the intended method when the base is "awkward".
- Always check whether a base can be rewritten as a power (for example $8=2^3$) before changing base, because that can simplify further.
