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Section 6.2 Properties of Logarithms, Solving Equations Using Logarithms

Algebraic properties of logarithms are what make them useful in applications. These properties are similar for all logarithms base \(b\text{.}\) For the sake of clarity, though, we will list and practice separately the properties of the common logarithm and the properties of the natural logarithm.

Subsection Algebraic Properties of Logarithms

We begin with the common logarithm.
Properties 1–2 follow from the definition of the logarithm base \(10\text{.}\) For Property 3, justification requires explaining why the logarithm of a product equals a sum of logarithms. From the definition of \(\log x = \log_{10} x\text{,}\) then for any number (or expression) \(w\text{,}\)
\begin{equation*} \log (A\cdot B) = w \quad \Leftrightarrow \quad 10^w = A\cdot B. \end{equation*}
Notice that from properties of exponential expressions,
\begin{equation*} 10^{\log A + \log B} = 10^{\log A}\cdot 10^{\log B}. \end{equation*}
Indeed, the sum of exponents corresponds to the product of the exponential expressions. From Property 2 of the common logarithm,
\begin{equation*} 10^{\log A} = A, \qquad 10^{\log B} = B. \end{equation*}
It follows that
\begin{equation*} 10^{\log A + \log B} = 10^{\log A}\cdot 10^{\log B} = A\cdot B. \end{equation*}
Therefore,
\begin{equation*} \log (A\cdot B) = \log A + \log B \quad \text{as} \quad 10^{\log A + \log B} = A\cdot B. \end{equation*}
Similarly, Properties 4–5 of the logarithm follow from the algebraic properties of exponential expressions.
All logarithms have similar properties as those listed above for the common logarithm. In particular, similar properties hold for the natural logarithm.

Subsection Solving Equations Using Logarithms

Property 5 of both logarithms is especially important and makes logarithms useful for solving equations in which the unknown is in the exponent.

Example 6.2.3.

Solve the equation:
\begin{equation*} 3^x = 600. \end{equation*}
Solution.
The unknown \(x\) is in the exponent, so we apply a logarithm to both sides of the equation. Either the common logarithm or the natural logarithm will work equally well, but we use the common logarithm. Taking the logarithm of both sides of \(3^x = 600\) gives us an equivalent equation:
\begin{equation*} \log(3^x) = \log(600). \end{equation*}
Note: We took the logarithm of each side; which is NOT to be interpreted as “multiplying each side by \(\log\)”. Similarly, we cannot multiply both sides of an equation by the radical “\(\sqrt{\quad}\)”. We can, however, take the radical of both sides.
Property 5 of the common logarithm takes the exponent out of the logarithm:
\begin{equation*} x\log(3) = \log(600). \end{equation*}
Note that \(\log(3)\) and \(\log(600)\) are just constants—you can calculate their approximate values using your calculator. The equation is then \(x\) times a constant is equal to another constant. Divide both sides of the equation by \(\log(3)\) to get the answer:
\begin{equation*} x = \frac{\log(600)}{\log(3)} \approx 5.823. \end{equation*}
You can check for yourself that you will get the same answer if you choose to take the natural logarithm of both sides rather than the common logarithm. The exact answers may look different, but the decimal approximation will show they are in fact the same.

Example 6.2.4.

Solve for \(t\text{:}\)
\begin{equation*} 1000\cdot 1.05^t = 2000. \end{equation*}
Solution.
Before applying the logarithm to both sides, it is best to divide both sides by \(1000\text{:}\)
\begin{equation*} 1.05^t = \frac{2000}{1000}. \end{equation*}
That is, \(1.05^t = 2\text{.}\) Now, apply the logarithm to both sides and solve:
\begin{align*} \log(1.05^t) \amp = \log(2)\\ t\log(1.05) \amp = \log(2) \amp\amp (\href{sec_6_2.html#com-log-prop-5}{\text{Property 5}})\\ \frac{t \, \cancel{\log(1.05)}}{\cancel{\log(1.05)}} \amp = \frac{\log(2)}{\log (1.05)}\\ t \amp = \frac{\log(2)}{\log(1.05)} \approx 14.207. \end{align*}
Notice that we also found the doubling time of the function \(y = 1000\cdot 1.05^t\) with this calculation.

Example 6.2.5.

Solve for \(t\text{:}\)
\begin{equation*} 500e^{0.072t} = 1800 \end{equation*}
Solution.
After dividing both sides by \(500\text{,}\)
\begin{equation*} e^{0.072t} = \frac{18}{5}. \end{equation*}
Since the base of the exponential expression is \(e\text{,}\) it will be a bit easier to use the natural logarithm. Applying the natural logarithm to both sides gives
\begin{equation*} \ln(e^{0.072t}) = \ln\left( \frac{18}{5} \right). \end{equation*}
There are two options at this time. The first is to use Property 5 of the natural logarithm with \(p = 0.072t\) and get:
\begin{equation*} \ln(e^{0.072t}) = 0.072t\ln(e) = 0.072t \quad \text{as} \quad \ln(e) = 1. \end{equation*}
Or, use Property 1, which says in essence that for any number or expression \(w\text{,}\) \(\ln(e^{w}) = w\text{.}\) Either way,
\begin{equation*} \ln(e^{0.072t}) = 0.072t \end{equation*}
and so
\begin{equation*} 0.072t = \ln \left(\frac{18}{5}\right). \end{equation*}
Dividing both sides by \(0.072\text{,}\) the exact and approximate solutions are
\begin{equation*} t = \frac{\ln(\frac{18}{5})}{0.072} \approx 17.79 \end{equation*}

Example 6.2.6.

Solve for \(x\) in the following equation. Give the exact answer as well as a decimal approximation.
\begin{equation*} 3\cdot 8^x = 5\cdot 6^x. \end{equation*}
Solution.
The unknown \(x\) appears in two exponential expressions. One approach is to combine those expressions into one by first dividing both sides of the equation by \(3\cdot 6^x\) which will result in
\begin{equation*} \frac{8^x}{6^x} = \frac{5}{3}. \end{equation*}
Now remember that the power of a quotient is the quotient of the corresponding powers: \(\dfrac{a^x}{b^x} = \left(\dfrac{a}{b}\right)^x\text{.}\) Hence, the equation becomes:
\begin{equation*} \left(\frac{8}{6}\right)^x = \frac{5}{3}. \end{equation*}
Now, apply the common logarithm to both sides:
\begin{equation*} \log\left(\left(\frac{8}{6}\right)^x\right) = \log\left(\frac{5}{3}\right). \end{equation*}
By Property 5, the left-hand side simplifies to the following:
\begin{equation*} x\log\left(\frac{8}{6}\right) = \log\left(\frac{5}{3}\right). \end{equation*}
Divide both sides by the constant \(\log\left(\frac{8}{6}\right)\text{,}\) yielding the answer:
\begin{equation*} x = \dfrac{\log\left(\frac{5}{3}\right)}{\log\left(\frac{8}{6}\right)} \approx 1.776. \end{equation*}
What would happen if we didn’t simplify the equation \(3\cdot 8^x = 5\cdot 6^x\) before applying the logarithm to both sides? Nothing much, except that the calculations would become a bit more complicated, since we would have
\begin{equation*} \log (3\cdot 8^x) = \log(5\cdot 6^x). \end{equation*}
This requires Property 3 to expand the logarithms on both sides as in:
\begin{equation*} \log(3) + \log (8^x) = \log(5) + \log (6^x). \end{equation*}
With Property 5, the exponent \(x\) can be pulled out and then
\begin{equation*} \log(3) + x\log( 8) = \log(5) + x\log (6). \end{equation*}
Notice that \(\log (3)\text{,}\) \(\log (5)\text{,}\) \(\log (6)\text{,}\) \(\log (8)\) are all constants and can be grouped on one side of the equation, and all terms involving \(x\) appear on the other side:
\begin{equation*} x(\log(8) - \log(6)) = \log(5) - \log(3). \end{equation*}
Simplifying, the answer is
\begin{equation*} x = \frac{\log(5) - \log(3)}{\log(8) - \log(6)} = \dfrac{\log\left(\frac{5}{3}\right)}{\log\left(\frac{8}{6}\right)}. \end{equation*}
Indeed, the difference of logarithms corresponds to the logarithm of a quotient by Property 4 . Hence, we obtained the same answer as before, but getting there required more steps.

Example 6.2.7.

Rewrite the following expressions in terms of \(\log x\) and \(\log y\text{,}\) or state that this is not possible.
  1. \(\displaystyle \log\left(\dfrac{5y}{x}\right)\)
  2. \(\displaystyle \log(x^2 - 5y)\)
  3. \(\displaystyle \ln\left(7x^3y\right)\)
Solution.
  1. Property 4 —the logarithm of a quotient is the difference of logarithms—to obtain
    \begin{align*} \log\left(\dfrac{5y}{x}\right) \amp= \log(5y) - \log(x)\\ \amp= \log(5) + \log(y) - \log(x) \end{align*}
  2. Before we can do anything else, we have to deal with the logarithm of a difference. There is no formula for the logarithm of a sum or a difference! This expression cannot be simplified.
  3. Using the properties listed above,
    \begin{equation*} \ln\left(7x^3y\right) = \ln(7) + \ln(x^3) + \ln(y) = \ln(7) + 3\ln(x) + \ln(y). \end{equation*}

Example 6.2.8.

Condense the following expression into one logarithm:
\begin{equation*} 3\ln y + \frac{1}{2}\ln(x) - \ln z \end{equation*}
Solution.
Using properties of the natural logarithm,
\begin{equation*} 3\ln y + \frac{1}{2}\ln(x) - \ln z = \ln (y^3) + \ln (x^{\frac{1}{2}}) + \ln (z^{-1}) = \ln\left(\frac{y^3\sqrt{x}}{z}\right). \end{equation*}

Subsection The Change of Base Formula for Logarithms

Even though many scientific and graphing calculators are only programmed to work with the common logarithm and the natural logarithm, you can calculate the value of any logarithm \(\log_b x\) thanks to the following change of base formulas:
\begin{equation*} \log_b x = \frac{\ln x}{\ln b}, \qquad \log_b x = \frac{\log x}{\log b} \end{equation*}

Example 6.2.9.

Use your calculator to find \(\log_2 5\text{.}\)
Solution.
Either change of base formula can be utilized here. Using the first one would give us
\begin{equation*} \log_2 5 = \frac{\ln 5}{\ln 2} \approx 2.322, \end{equation*}
rounded to three decimal places. The answer is the same if we convert to the common logarithm:
\begin{equation*} \log_2 5 = \frac{\log 5}{\log 2} \approx 2.322. \end{equation*}

Exercises Exercises

Properties of Logarithms.

For each of the following, use properties of logarithms to expand a given expression as much as possible and write it in terms of \(\log(x)\text{,}\) \(\log(y)\) and \(\log(z)\text{,}\) if possible. Assume that \(x\text{,}\) \(y\text{,}\) and \(z\) are positive.

1.

\(\displaystyle \log\left(\frac{5z}{x}\right)\)
Solution.
\(\log(5)+\log(z)-\log(x)\)

2.

\(\displaystyle \log(x^5y^3z^2)\)
Solution.
\(5\log(x)+3\log(y)+2\log(z)\)

3.

\(\displaystyle \log\left(\frac{yz}{6x}\right)\)
Solution.
\(\log(y)+\log(z)-\log(6)-\log(x)\)

4.

\(\displaystyle \log\left(\frac{y^2z^3}{\sqrt{4x}}\right)\)
Solution.
\(2\log(y)+3\log(z)-\frac{1}{2}\log(4)-\frac{1}{2}\log(x)\)

5.

\(\displaystyle \log\left(\frac{y^2z}{x}-x^2\right)\)
Solution.
Cannot simplify.

6.

\(\displaystyle \log(x^2+y^2)\)
Solution.
Cannot simplify.

7.

\(\displaystyle \log(100y\sqrt{x})\)
Solution.
\(2+\log(y)+\frac{1}{2}\log(x)\)

8.

\(\displaystyle \log\left(\frac{10x}{zy}\right)\)
Solution.
\(1+\log(x)-\log(z)-\log(y)\)

Expanding Logarithms.

For each of the following, use properties of logarithms to expand a given expression as much as possible and write it in terms of \(\ln(x)\text{,}\) \(\ln(y)\) and \(\ln(z)\text{,}\) if possible. Assume that \(x\text{,}\) \(y\text{,}\) and \(z\) are positive.

9.

\(\displaystyle \ln\left(\frac{z^2}{2x}\right)\)
Solution.
\(2\ln(z)-\ln(2)-\ln(x)\)

10.

\(\displaystyle \ln\left( \sqrt{(x^3y^3z)} \right)\)
Solution.
\(\frac{3}{2}\ln(x)+\frac{3}{2}\ln(y)+\frac{1}{2}\ln(z)\)

12.

\(\displaystyle \ln\left(\frac{exy}{3z}\right)\)
Solution.
\(1+\ln(x)+\ln(y)-\ln(3)-\ln(z)\)

13.

\(\displaystyle \ln(e^2yz^{-3})\)
Solution.
\(2+\ln(y)-3\ln(z)\)

14.

\(\displaystyle \ln\left(\frac{x}{3e}\right)\)
Solution.
\(\ln(x)-\ln(3)-1\)

15.

\(\displaystyle \ln\left(\frac{1}{x^3y}\right)\)
Solution.
\(-3\ln(x)-\ln(y)\)

16.

\(\displaystyle \ln\left(\frac{\sqrt{e}}{xz}\right)\)
Solution.
\(\frac{1}{2}-\ln(x)-\ln(z)\)

Simplifying Logarithms.

For each of the following, combine the expression into one logarithm, if possible. Assume that \(A\text{,}\) \(B\text{,}\) and \(C\) are positive.

17.

\(\displaystyle \frac{\log(A)}{\log(B)}\)
Solution.
Not possible.

18.

\(\displaystyle 2\log(A)-3\log(B)\)
Solution.
\(\log\left(\frac{A^2}{B^3}\right)\)

19.

\(\displaystyle -3\ln(B)+\frac{1}{2}\ln(AB)\)
Solution.
\(\ln\left(B^{-3}\sqrt{AB}\right)\)

20.

\(\displaystyle \log(A)\cdot \log(B)\)
Solution.
Not possible.

21.

\(\displaystyle \frac{1}{3}\ln(A)+2\ln(B)-\ln(10)\)
Solution.
\(\ln\left(\frac{\sqrt[3]{A}B^2}{10}\right)\)

22.

\(\displaystyle \log(5)\cdot\log(AB)\)
Solution.
Not possible.

Solving Exponential Equations.

For each of the following, solve a given equation for \(x\) or for \(t\text{.}\) Give the exact answer as well as its approximation rounded off to three decimal places.

23.

\(\displaystyle 5^x=500\)
Solution.
\(x=\dfrac{\log(500)}{\log(5)}\approx3.861\)

24.

\(\displaystyle 1200\cdot 1.15^t=2400\)
Solution.
\(t=\dfrac{\log(2)}{\log(1.15)}\approx 4.959\)

25.

\(\displaystyle 250\cdot 0.81^t=125\)
Solution.
\(t=\dfrac{\log\left(\frac{1}{2}\right)}{\log(0.81)}\approx3.289\)

26.

\(\displaystyle 3\cdot 7^x=4\cdot 5^x\)
Solution.
\(x=\dfrac{\log\left(\frac{4}{3}\right)}{\log\left(\frac{7}{5}\right)}\approx0.855\)

27.

\(\displaystyle 90(0.5)^{t/10}=40\)
Solution.
\(t=\dfrac{10\log\left(\frac{4}{9}\right)}{\log(0.5)}\approx11.69\)

28.

\(\displaystyle 490e^{-0.098t}=100\)
Solution.
\(t=-\dfrac{\ln\left(\frac{10}{49}\right)}{0.098}\approx16.217\)

Using the Change of Base Formula.

For each of the following, rewrite a given logarithm in terms of the natural logarithm and calculate its value using your calculator. Round off to three decimal places.

29.

\(\displaystyle \log_5 (8)\)
Solution.
\(\dfrac{\ln(8)}{\ln(5)}\approx1.292\)

30.

\(\displaystyle \log_3 (12)\)
Solution.
\(\dfrac{\ln(12)}{\ln(3)}\approx2.262\)
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