Using Exponent and Logarithm Laws, foundations for calculus lesson

A product like $e^{2x} \cdot e^{3 - x}$ or a logarithm like $\ln\left(x^2 \sqrt{y}\right)$ is awkward to work with as written. The exponent and logarithm laws let you turn it into a single power or a sum of simple pieces, and calculus asks for exactly that before it differentiates or integrates. Each law also comes with conditions, and a rewrite that ignores them can quietly change where the expression is defined.

Key terms

Product of powers

Definition For a base $a > 0$ and any real exponents $p$ and $q$ ,

a^p \cdot a^q = a^{p + q}.

When $p$ and $q$ are integers the law also holds for any base $a \ne 0$ , but with real exponents the base must be positive, because powers like $a^{1/2}$ are not real numbers for $a < 0$ .

In plain words Two powers of the same base multiply by adding their exponents. The base has to be the same on both factors.

Example Suppose you want to combine $e^{2x} \cdot e^{3 - x}$ into one power. The base $e$ is positive and both factors share it, so add the exponents.

e^{2x} \cdot e^{3 - x} = e^{2x + 3 - x} = e^{x + 3}.

At $x = 1$ the left side is $e^2 \cdot e^2 = e^4$ and the right side is $e^4$ . The rewrite holds for every real $x$ , so nothing about the domain changed. By contrast $2^3 \cdot 5^3$ has two different bases and is not $10^6$ . It equals $1000 = 10^3$ .

Quotient of powers

Definition For a base $a > 0$ and any real exponents $p$ and $q$ ,

\frac{a^p}{a^q} = a^{p - q}.

The denominator is never zero, since $a^q > 0$ for every real $q$ when $a > 0$ .

In plain words Dividing powers of the same base subtracts the exponent of the bottom from the exponent of the top.

Example Here the variable disappears completely.

\frac{3^{x + 2}}{3^{x - 1}} = 3^{(x + 2) - (x - 1)} = 3^3 = 27.

So the ratio is the constant $27$ for every real $x$ . A quick check at $x = 0$ gives $9 \div \frac{1}{3} = 27$ .

Power of a power

Definition For a base $a > 0$ and any real exponents $p$ and $q$ ,

\left(a^p\right)^q = a^{pq}.

The condition $a > 0$ matters. With a negative base and a fractional exponent the law can fail. For example $\left((-1)^2\right)^{1/2} = 1^{1/2} = 1$ , while $(-1)^{2 \cdot \frac{1}{2}} = (-1)^1 = -1$ .

In plain words Raising a power to another power multiplies the exponents, as long as the base is positive.

Example Rewriting $\left(2^x\right)^3$ as a single exponential with base $2$ , and then with exponent $x$ alone, gives

\left(2^x\right)^3 = 2^{3x} = \left(2^3\right)^x = 8^x.

At $x = 2$ the left side is $4^3 = 64$ and the right side is $8^2 = 64$ .

Logarithm product law

Definition Let $a > 0$ with $a \ne 1$ . For all $u > 0$ and $v > 0$ ,

\log_a(uv) = \log_a u + \log_a v.

It comes from the product of powers. If $u = a^m$ and $v = a^n$ , then $uv = a^{m + n}$ , so the logarithm of $uv$ is $m + n$ .

In plain words The logarithm of a product is the sum of the logarithms, provided each factor is positive.

Example With base $2$ ,

\log_2(8 \cdot 32) = \log_2 256 = 8,

and on the other side $\log_2 8 + \log_2 32 = 3 + 5 = 8$ . With a variable, $\ln(3x) = \ln 3 + \ln x$ holds for $x > 0$ , which is exactly the domain of $\ln(3x)$ .

Logarithm quotient law

Definition Let $a > 0$ with $a \ne 1$ . For all $u > 0$ and $v > 0$ ,

\log_a\left(\frac{u}{v}\right) = \log_a u - \log_a v.

In plain words The logarithm of a quotient is the logarithm of the top minus the logarithm of the bottom.

Example With base $10$ ,

\log_{10}\left(\frac{1000}{0.1}\right) = \log_{10} 10000 = 4,

and $\log_{10} 1000 - \log_{10} 0.1 = 3 - (-1) = 4$ . A logarithm can be negative, as $\log_{10} 0.1$ is, even though its argument must be positive.

Logarithm power law

Definition Let $a > 0$ with $a \ne 1$ . For every $u > 0$ and every real number $p$ ,

\log_a\left(u^p\right) = p \log_a u.

The condition $u > 0$ is what makes $\log_a u$ on the right side defined. Dropping it changes the domain. For example $\ln\left(x^2\right)$ is defined for every $x \ne 0$ , but $2 \ln x$ is defined only for $x > 0$ . For $x \ne 0$ the correct statement is $\ln\left(x^2\right) = 2 \ln \lvert x \rvert$ .

In plain words An exponent inside a logarithm can come out in front as a factor, provided the quantity being raised is positive.

Example Since $8 = 2^3$ ,

\log_2\left(8^5\right) = 5 \log_2 8 = 5 \cdot 3 = 15,

which matches $8^5 = 2^{15}$ . Roots work too, because a root is a power. For $x > 0$ you have $\ln \sqrt{x} = \frac{1}{2} \ln x$ .

Change of base

Definition Let $a$ and $b$ be valid bases, so $a > 0$ , $a \ne 1$ , $b > 0$ and $b \ne 1$ . For every $x > 0$ ,

\log_a x = \frac{\log_b x}{\log_b a}.

The denominator is not zero, because $\log_b a = 0$ would mean $a = 1$ .

In plain words Any logarithm can be written with a base you prefer, usually $e$ or $10$ , by dividing two logarithms in that base.

Example A calculator has $\ln$ but no base $2$ key. To find $\log_2 10$ ,

\log_2 10 = \frac{\ln 10}{\ln 2} \approx 3.3219.

The sign $\approx$ marks a rounded value. Sometimes the answer is exact. Using base $2$ ,

\log_8 32 = \frac{\log_2 32}{\log_2 8} = \frac{5}{3}.

Positive arguments and preserving the domain

Definition Every logarithm law above is an identity only where all the logarithms in it are defined, so every argument must be positive. A rewrite preserves the domain when the new form is defined at exactly the same inputs as the original and has the same value at each of them. If the new form is defined on a smaller or larger set, the two forms are equal only where both are defined, and that set must be stated next to the rewrite. There is no law for the logarithm of a sum. In general

\log_a(u + v) \ne \log_a u + \log_a v.

The correct product law involves $\log_a(uv)$ , not $\log_a(u + v)$ .

In plain words Before splitting a logarithm, ask where each piece is defined. If the pieces live on a smaller set than the original, the split is only true on that smaller set, and you should say so.

Example A sum inside a logarithm does not split. With $u = v = 1$ the left side is $\ln 2 \approx 0.693$ and the right side is $\ln 1 + \ln 1 = 0$ .

Domain can shrink too. The expression $\ln\left(x(x - 1)\right)$ is defined when $x(x - 1) > 0$ , which means $x < 0$ or $x > 1$ . The split form $\ln x + \ln(x - 1)$ needs $x > 0$ and $x > 1$ , so it lives only on $x > 1$ . At $x = -1$ the original equals $\ln 2$ while the split form is undefined. So write

\begin{gathered} \ln\left(x(x - 1)\right) = \ln x + \ln(x - 1) \\[0.4em] \text{for } x > 1. \end{gathered}

When positivity is assumed from the start, say it first. For $x > 0$ , $y > 0$ and $z > 0$ ,

\begin{aligned} \ln\left(\frac{x^2 \sqrt{y}}{z^3}\right) &= 2 \ln x + \frac{1}{2} \ln y \\[0.4em] &\quad - 3 \ln z. \end{aligned}

The quotient law split off $\ln\left(z^3\right)$ , the product law split $\ln\left(x^2 \sqrt{y}\right)$ , and the power law brought each exponent forward.

Check yourself

Write $\dfrac{e^{3x} \cdot e^{-x}}{e^{x + 2}}$ as a single power of $e$ .

Show the answer

Add the exponents in the numerator and subtract the exponent of the denominator.

\frac{e^{3x} \cdot e^{-x}}{e^{x + 2}} = e^{3x - x - (x + 2)} = e^{x - 2}.

Both forms are defined for every real $x$ .

Assume $x > 0$ , $y > 0$ and $z > 0$ . Expand $\ln\left(\dfrac{x^4}{y \sqrt{z}}\right)$ into a sum and difference of simple logarithms.

Show the answer

The quotient law gives $\ln\left(x^4\right) - \ln\left(y \sqrt{z}\right)$ , the product law splits the second term, and the power law finishes.

\ln\left(\frac{x^4}{y \sqrt{z}}\right) = 4 \ln x - \ln y - \frac{1}{2} \ln z.

Is $\ln(x - 2) + \ln(x + 2) = \ln\left(x^2 - 4\right)$ true at every $x$ where the right side is defined?

Show the answer

No. The right side is defined for $x < -2$ or $x > 2$ , but the left side needs $x > 2$ . At $x = -3$ the right side is $\ln 5$ and the left side is undefined. The equation holds exactly for $x > 2$ .

Using Exponent and Logarithm Laws

Key terms

Product of powers

Quotient of powers

Power of a power

Logarithm product law

Logarithm quotient law

Logarithm power law

Change of base

Positive arguments and preserving the domain

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus