14.1: Algebraic Properties of exp and log

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    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. \(\begin{aligned}
      \ln \left(\sqrt{x^{5}} \cdot y^{2}\right)&=\ln \left(x^{\frac{5}{2}} \cdot y^{2}\right)\\
      &=\ln \left(x^{\frac{5}{2}}\right)+\ln \left(y^{2}\right) \\
      &=\dfrac{5}{2} \ln (x)+2 \ln (y)\\
      &=\dfrac{5}{2} u+2 v
      \end{aligned}\)
    1. \(\begin{aligned}
      \log \left(\sqrt{\sqrt{x} \cdot y^{3}}\right)&=\log \left(\left(x^{\frac{1}{2}} y^{3}\right)^{\frac{1}{2}}\right)\\
      &=\dfrac{1}{2} \log \left(x^{\frac{1}{2}} y^{3}\right) \\
      &=\dfrac{1}{2}\left(\log \left(x^{\frac{1}{2}}\right)+\log \left(y^{3}\right)\right)\\
      &=\dfrac{1}{2}\left(\dfrac{1}{2} \log (x)+3 \log (y)\right) \\
      &=\dfrac{1}{4} \log (x)+\dfrac{3}{2} \log (y)\\
      &=\dfrac{1}{4} u+\dfrac{3}{2} v
      \end{aligned}\)
    1. \(\begin{aligned}\log _{2}\left(\sqrt[3]{\dfrac{x^{2}}{y \sqrt{z}}}\right) &=\log _{2}\left(\left(\dfrac{x^{2}}{y \cdot z^{\frac{1}{2}}}\right)^{\frac{1}{3}}\right)=\dfrac{1}{3} \log _{2}\left(\dfrac{x^{2}}{y \cdot z^{\frac{1}{2}}}\right) \\ &=\dfrac{1}{3}\left(\log _{2}\left(x^{2}\right)-\log _{2}(y)-\log _{2}\left(z^{\frac{1}{2}}\right)\right) \\ &=\dfrac{1}{3}\left(2 \log _{2}(x)-\log _{2}(y)-\dfrac{1}{2} \log _{2}(z)\right) \\ &=\dfrac{2}{3} \log _{2}(x)-\dfrac{1}{3} \log _{2}(y)-\dfrac{1}{6} \log _{2}(z) \\ &=\dfrac{2}{3} u-\dfrac{1}{3} v-\dfrac{1}{6} w \end{aligned}\)

    This page titled 14.1: Algebraic Properties of exp and log was last modified on Mon, 02 May 2022 15:49:53 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform.