14.1: Algebraic Properties of exp and log

We recall some of the main properties of the exponential function:

\[\boxed{\begin{matrix} \quad b^{x + y} = b^x\cdot b^y\quad \\ b^{x-y} = \dfrac{b^x}{b^y} \\ (b^x)^n = b^{nx} \end{matrix}} \nonumber \]

Writing the above properties in terms of \(f(x)=b^x\) we have \(f(x+y)=f(x)f(y)\), \(f(x-y)=f(x)/f(y)\), and \(f(nx)=f(x)^n\).

Due to the fact that the logarithm is the inverse function of the exponential, we have corresponding properties whose proof follows:

Proposition: Logarithm Properties

The logarithm behaves well with respect to products, quotients, and exponentiation. Indeed, for all positive real numbers \(0<b\neq 1\), \(x>0\), \(y>0\), and real numbers \(n\), we have:

\[\boxed{\begin{matrix} \quad\log_b(x\cdot y) = \log_b(x)+\log_b(y)\quad \\ \log_b(\dfrac{x}{y}) = \log_b(x)-\log_b(y) \\ \log_b(x^n) = n\cdot \log_b(x) \end{matrix}} \nonumber \]

In terms of the logarithmic function \(g(x)=\log_b(x)\), the properties in the table above can be written: \(g(xy)=g(x)+g(y)\), \(g(x/y)=g(x)-g(y)\), and \(g(x^n)=n\cdot g(x)\).

Furthermore, for another positive real number \(0<a\neq 1\), we have the change of base formula:

\[\label{EQU:log-change-base} \boxed{\quad\log_b(x)=\dfrac{\log_a(x)}{\log_a(b)}\quad} \]

In particular, we have the formulas from equation 13.2.2 when taking the base \(a=10\) and \(a=e\):

\[\log_b(x) = \dfrac{\log(x)}{\log(b)}\quad\quad\text{and}\quad\quad \log_b(x) = \dfrac{\ln(x)}{\ln(b)} \nonumber \]

Proof

We start with the first formula \(\log_b(x\cdot y) = \log_b(x)+\log_b(y)\). If we call \(u=\log_b(x)\) and \(v=\log_b(y)\), then the equivalent exponential formulas are \(b^u=x\) and \(b^v=y\). With this, we have

\[x\cdot y = b^u\cdot b^v =b^{u+v} \nonumber \]

Rewriting this in logarithmic form, we obtain

\[\log_b(x\cdot y)=u+v=\log_b(x)+\log_b(y) \nonumber \]

This is what we needed to show.

Next, we prove the formula \(\log_b\left(\dfrac{x}{y}\right) = \log_b(x)-\log_b(y)\). We abbreviate \(u=\log_b(x)\) and \(v=\log_b(y)\) as before, and their exponential forms are \(b^u=x\) and \(b^v=y\). Therefore, we have

\[\dfrac{x}{y} = \dfrac{b^u}{b^v} =b^{u-v} \nonumber \]

Rewriting this again in logarithmic form, we obtain the desired result.

\[\log_b\Big(\dfrac{x}{y}\Big)=u-v=\log_b(x)-\log_b(y) \nonumber \]

For the third formula, \(\log_b(x^n) = n\cdot \log_b(x)\), we write \(u=\log_b(x)\), that is in exponential form \(b^u=x\). Then:

\[x^n=(b^u)^n=b^{n\cdot u} \quad \implies \quad \log_b(x^n)= n\cdot u =n\cdot \log_b(x) \nonumber \]

For the last formula \(\ref{EQU:log-change-base}\), we write \(u=\log_b(x)\), that is, \(b^u=x\). Applying the logarithm with base \(a\) to \(b^u=x\) gives \(\log_a(b^u)=\log_a(x)\). As we have just shown before, \(\log_a(b^u)=u\cdot \log_a(b)\). Combining these identities with the initial definition \(u=\log_b(x)\), we obtain

\[\log_a(x)= \log_a(b^u)= u\cdot \log_a(b) =\log_b(x)\cdot \log_a(b) \nonumber \]

Dividing both sides by \(\log_a(b)\) gives the result \(\dfrac{\log_a(x)}{\log_a(b)}=\log_b(x)\).

Example \(\PageIndex{1}\)

Combine the terms using the properties of logarithms so as to write as one logarithm.

1. \(\dfrac{1}{2}\ln(x)+\ln(y)\)
2. \(\dfrac{2}{3}(\log(x^2y)-\log(xy^2))\)
3. \(2\ln(x)-\dfrac{1}{3}\ln(y)-\dfrac{7}{5}\ln(z)\)
4. \(5+\log_2(a^2-b^2)-\log_2(a+b)\)

Solution

Recall that a fractional exponent can also be rewritten with an \(n\)th root.

\[\boxed{x^{\frac 1 2} = \sqrt x} \quad\text{ and }\quad \boxed{x^{\frac 1 n} = \sqrt[n]x} \quad\implies \quad x^{\frac p q}=(x^p)^{\frac{1} {q}}=\sqrt[q]{x^p} \nonumber \]

We apply the rules from proposition Logarithm Properties.

1. \(\dfrac{1}{2}\ln(x)+\ln(y)= \ln(x^{\frac 1 2})+\ln(y)=\ln(x^{\frac 1 2} y)=\ln(\sqrt{x}\cdot y)\)
2. \(\dfrac{2}{3}(\log(x^2y)-\log(xy^2))=\dfrac{2}{3}\left(\log\left(\dfrac{x^2y}{xy^2}\right)\right)=\dfrac{2}{3}\left(\log\left(\dfrac{x}{y}\right)\right)\) &\(=\log\left(\left(\dfrac{x}{y}\right)^{\frac{2}{3}}\right)=\log\left(\sqrt[3]{\dfrac {x^2} {y^2}}\,\,\right)\)
3. \(2\ln(x)-\dfrac{1}{3}\ln(y)-\dfrac{7}{5}\ln(z)=\ln(x^2)-\ln(\sqrt[3]{y})-\ln(\sqrt[5]{z^7})=\ln \left(\dfrac{x^2}{\sqrt[3]{y}\cdot \sqrt[5]{z^7}}\right)\)
4. \(5+\log_2(a^2-b^2)-\log_2(a+b)=\log_2(2^5)+\log_2(a^2-b^2)-\log_2(a+b)\) &\(=\log_2\left(\dfrac{2^5\cdot (a^2-b^2)}{a+b}\right)=\log_2\left(\dfrac{32\cdot (a+b)(a-b)}{a+b}\right)=\log_2(32\cdot(a-b))\)

Example \(\PageIndex{2}\)

Write the expressions in terms of elementary logarithms \(u=\log_b(x)\), \(v=\log_b(y)\), and, in part (c), also \(w=\log_b(z)\). Assume that \(x,y,z>0\).

1. \(\ln( \sqrt{x^5}\cdot y^2)\)
2. \(\log \left(\sqrt{\sqrt{x} \cdot y^{3}}\right)\)
3. \(\log _{2}\left(\sqrt[3]{\dfrac{x^{2}}{y \sqrt{z}}}\right)\)

Solution

In a first step, we rewrite the expression with fractional exponents, and then apply the rules from proposition Logarithm Properties.

1. \(\begin{aligned}
   \ln \left(\sqrt{x^{5}} \cdot y^{2}\right)&=\ln \left(x^{\frac{5}{2}} \cdot y^{2}\right)\\
   &=\ln \left(x^{\frac{5}{2}}\right)+\ln \left(y^{2}\right) \\
   &=\dfrac{5}{2} \ln (x)+2 \ln (y)\\
   &=\dfrac{5}{2} u+2 v
   \end{aligned}\)

2. \(\begin{aligned}
   \log \left(\sqrt{\sqrt{x} \cdot y^{3}}\right)&=\log \left(\left(x^{\frac{1}{2}} y^{3}\right)^{\frac{1}{2}}\right)\\
   &=\dfrac{1}{2} \log \left(x^{\frac{1}{2}} y^{3}\right) \\
   &=\dfrac{1}{2}\left(\log \left(x^{\frac{1}{2}}\right)+\log \left(y^{3}\right)\right)\\
   &=\dfrac{1}{2}\left(\dfrac{1}{2} \log (x)+3 \log (y)\right) \\
   &=\dfrac{1}{4} \log (x)+\dfrac{3}{2} \log (y)\\
   &=\dfrac{1}{4} u+\dfrac{3}{2} v
   \end{aligned}\)

3. \(\begin{aligned}\log _{2}\left(\sqrt[3]{\dfrac{x^{2}}{y \sqrt{z}}}\right) &=\log _{2}\left(\left(\dfrac{x^{2}}{y \cdot z^{\frac{1}{2}}}\right)^{\frac{1}{3}}\right)=\dfrac{1}{3} \log _{2}\left(\dfrac{x^{2}}{y \cdot z^{\frac{1}{2}}}\right) \\ &=\dfrac{1}{3}\left(\log _{2}\left(x^{2}\right)-\log _{2}(y)-\log _{2}\left(z^{\frac{1}{2}}\right)\right) \\ &=\dfrac{1}{3}\left(2 \log _{2}(x)-\log _{2}(y)-\dfrac{1}{2} \log _{2}(z)\right) \\ &=\dfrac{2}{3} \log _{2}(x)-\dfrac{1}{3} \log _{2}(y)-\dfrac{1}{6} \log _{2}(z) \\ &=\dfrac{2}{3} u-\dfrac{1}{3} v-\dfrac{1}{6} w \end{aligned}\)