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Using Properties of Logarithms

Rule

Properties of Logarithms

The Properties of Logarithms allow expressions with logarithms to be rewritten.

Rule

Product Property of Logarithms

Rule

logb(mn)=logb(m)+logb(n)\log_b(mn)=\log_b(m)+\log_b(n)
The logarithm of a product can be written as the sum of the logarithms of each factor in the product. For example, log(74)\log(7\cdot4) can be expressed using this rule. log(74)=log(7)+log(4) \log(7\cdot4)=\log(7)+\log(4) This rule can be explained using the identity x=blogb(x).x=b^{\log_b(x)}.
logb(mn)\log_b(mn)
logb(mn)\log_b(m\cdot n)
m=blogb(m)m={b}^{\log_{b}(m)}
logb(blogb(m)blogb(n))\log_b\left(b^{\log_b(m)}\cdot b^{\log_b(n)}\right)
logb(blogb(m)+logb(n))\log_b\left(b^{\log_b(m)+\log_b(n)}\right)
logb(bm)=m\log_{b}\left({b}^m\right)=m
logb(m)+logb(n)\log_b(m)+\log_b(n)
This rule is valid for positive values of b,b, m,m, and nn and b1.b\neq 1.
Rule

Quotient Property of Logarithms

Rule

logb(mn)=logb(m)logb(n)\log_b\left(\dfrac{m}{n}\right)=\log_b(m)-\log_b(n)
The logarithm of a quotient can be written as the difference between the logarithm of the numerator and the logarithm of the denominator. For example, log(74)\log\left(\frac{7}{4} \right) can be expressed using this rule. log(74)=log(7)log(4) \log\left(\dfrac{7}{4}\right)=\log(7)-\log(4) This rule can be explained using the identity x=blogb(x).x=b^{\log_b(x)}.
logb(mn)\log_b\left(\dfrac{m}{n} \right)
m=blogb(m)m={b}^{\log_{b}(m)}
logb(blogb(m)blogb(n))\log_b\left(\dfrac{b^{\log_b(m)}}{b^{\log_b(n)}}\right)
logb(blogb(m)logb(n))\log_b\left(b^{\log_b(m)-\log_b(n)}\right)
logb(bm)=m\log_{b}\left({b}^m\right)=m
logb(m)logb(n)\log_b(m)-\log_b(n)
This rule is valid for positive values of b,b, m,m, and nn and b1.b\neq 1.
Rule

Power Property of Logarithms

Rule

logb(mn)=nlogb(m)\log_b\left(m^n\right)=n\log_b(m)
The logarithm of a power can be written as the product of the exponent and the logarithm of the base. For example, log(74)\log \left(7^4 \right) can be expressed using this rule. log(74)=4log(7) \log \left(7^4 \right)=4\cdot\log(7) This rule can be explained using the identity x=blogb(x).x=b^{\log_b(x)}.
logb(mn)\log_b(m^n)
m=blogb(m)m={b}^{\log_{b}(m)}
logb((blogb(m))n)\log_b\left (\left(b^{\log_b(m)} \right)^n\right)
logb(blogb(m)n)\log_b\left (b^{\log_b(m)\cdot n}\right)
logb(bm)=m\log_{b}\left({b}^m\right)=m
logb(m)n\log_b(m)\cdot n
nlogb(m)n\log_b(m)
This rule is valid for positive values of b,b, m,m, and nn and b1.b\neq 1.
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Exercise

Simplify the expression log6(8)log6(3)+log6(9)log6(4) \log_6(8)-\log_6(3)+\log_6(9)-\log_6(4) using the properties of logarithms.

Show Solution
Solution
The Quotient Property of Logarithms states that the following is true for logarithms. logb(mn)=logb(m)logb(n) \log_b\left(\dfrac{m}{n}\right)=\log_b(m)-\log_b(n) In our expression the first two terms and the last two terms both are a difference of two logarithms with the same base. We can use the Quotient Property of Logarithms to rewrite each pair into a logarithm of a quotient. log6(8)log6(3)+log6(9)log6(4)log6(83)+log6(94) \log_6(8)-\log_6(3)+\log_6(9)-\log_6(4) \\ \Updownarrow \\ \log_6 \left(\dfrac{8}{3} \right)+\log_6 \left(\dfrac{9}{4} \right) This expression is a sum of two logarithms with the same base. We can write them into a logarithm of a product using the Product Property of Logarithms. log6(83)+log6(94)log6(8394) \log_6 \left(\dfrac{8}{3} \right)+\log_6 \left(\dfrac{9}{4} \right) \quad \Leftrightarrow \quad \log_6 \left(\dfrac{8}{3} \cdot \dfrac{9}{4} \right) Let's continue simplifying this expression.
log6(8394)\log_6 \left(\dfrac{8}{3} \cdot \dfrac{9}{4} \right)
log6(8934)\log_6 \left(\dfrac{8\cdot 9}{3\cdot4} \right)
log6(6)\log_6 \left(6 \right)
log6(6)=1 \log_{6}({6}) = 1
11
Thus, the expression log6(8)log6(3)+log6(9)log6(4)\log_6(8)-\log_6(3)+\log_6(9)-\log_6(4) equals 1.1.
Rule

Change of Base Formula

The Change of Base Formula allows the logarithm of an arbitrary base to be rewritten as the quotient of two logarithms with another base. logc(a)=logb(a)logb(c) \log_c (a)= \dfrac{\log_b (a)}{\log_b (c)} With many calculators it is only possible to evaluate the common and the natural logarithm. The Change of Base Formula can then be used to evaluate logarithms of other bases.

logc(a)=log(a)log(c)andlogc(a)=ln(a)ln(c) \log_c (a)= \frac{\log (a)}{\log (c)} \quad \text{and} \quad \log_c (a)= \frac{\ln (a)}{\ln (c)}
Method

Solving an Exponential Equation using Logarithms

An exponential equation can be solved by applying logarithms and using the Power Property of Logarithms. Consider the following equation. 8x=3\begin{gathered} 8^x=3 \end{gathered}

1

Apply the logarithm

First, the equation is rewritten by applying a logarithm on both sides. 8x=3log(8x)=log(3) 8^x=3 \quad \Leftrightarrow \quad \log\left( 8^x \right)=\log(3)

2

Rewrite using the Power Property of Logarithms

By using the Power Property of Logarithms, powers can be rewritten into a product. log(8x)=log(3)xlog(8)=log(3)\begin{gathered} \log\left( 8^x \right)=\log(3) \\ \Updownarrow \\ x\cdot \log(8)=\log(3) \end{gathered}

3

Solve the resulting equation

After the power has been rewritten into a product, the unknown variable can be isolated using inverse operations. Here, xx gets isolated on the left-hand side when both sides of the equation are divided by log(8).\log(8). xlog(8)=log(3)x=log(3)log(8)\begin{gathered} x\cdot \log(8)=\log(3) \\ \Updownarrow \\ x=\dfrac{\log(3)}{\log(8)} \end{gathered} By using a calculator, an approximate value of xx can be calculated. Here, x0.53.x\approx 0.53.

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Exercise

Solve the equation 4x=25 4^x=25 using the common logarithm. State the answer with three significant digits.

Show Solution
Solution
The given equation can be solved by applying the log4\log_4 on both sides. Since many calculators are limited to only the common and the natural logarithms, it is often not possible to use log4.\log_4. We can solve it anyway. Let's begin by applying the common logarithm on both sides. 4x=25log(4x)=log(25) 4^x=25 \quad \Leftrightarrow \quad \log\left( 4^x \right)=\log(25) The Power Property of Logarithms gives us the relationship logb(mn)=nlogb(m).\log_b\left(m^n \right)=n\cdot \log_b(m). With this we can rewrite the left-hand side of the equation. log(4x)=log(25)xlog(4)=log(25) \log\left( 4^x \right)=\log(25)\, \, \Leftrightarrow \, \, x\cdot \log(4)=\log(25) We can now solve the equation for xx by dividing both sides of the equation with log(4).\log(4).
xlog(4)=log(25)x\cdot \log(4)=\log(25)
x=log(25)log(4)x=\dfrac{\log(25)}{\log(4)}
x=2.32192x=2.32192\ldots
x2.32x \approx 2.32
The solution to the equation is x2.32.x\approx 2.32.
Rule

Inverse Properties of Logarithms

A logarithmic function g(x)=logb(x),g(x)=\log_b (x), is by definition the inverse of an exponential function f(x)=bx.f(x)=b^x. This means that their function composition results in the identity function.

  1. g(f(x))=logb(bx)=xg(f(x))=\log_b \left( b^x \right) = x
  2. f(g(x))=blogb(x)=xf(g(x))=b^{\log _b (x)} = x

The fact that a logarithm and a power with the same base undo each other is what is known as the inverse properties of logarithms.

logb(bx)=xandblogb(x)=x\begin{gathered} \log_b\left(b^x\right)=x \quad \text{and} \quad b^{\log_b(x)}=x \end{gathered}

They also hold true for the common logarithm and the natural logarithm.

log(10x)=xand10log(x)=xln(ex)=xandeln(x)=x\begin{aligned} \log\left(10^x\right)=x \quad \text{and}& \quad 10^{\log(x)}=x \\[0.8em] \ln\left(e^x\right)=x \quad \text{and}& \quad e^{\ln(x)}=x \end{aligned}

These properties together with other properties of logarithms permit to simplify logarithmic expressions and to solve equations involving logarithms and powers. Some particular examples are shown below.

ln(e4x)=204x=20x=53log3(5x)=105x=10x=2\begin{array}{c|c} \begin{aligned} {\color{#0000FF}{\ln\left( e^{4x}\right)}}&=20\\ {\color{#0000FF}{4x}}&=20\\ x &= 5 \end{aligned} \hspace{0.35cm} & \hspace{0.35cm} \begin{aligned} {\color{#0000FF}{3^{\log_3 (5x)}}}&=10\\ {\color{#0000FF}{5x}}&=10\\ x &= 2 \end{aligned} \end{array}

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Exercise

Solve the equation eln(34)10log(x)log5(125)=log(100) \dfrac{e^{\ln (3^4)}\cdot 10^{\log(x)}}{\log_5\left( 125 \right)}=\log(100) using the inverse properties of logarithms.

Show Solution
Solution
The inverse properties of logarithms tell us that a logarithm and a power with the same base undo each other. logb(bx)=xandblogb(x)=x \log_b(b^x)=x \quad \text{and} \quad b^{\log_b(x)}=x On the left-hand side, we can identify ln\ln as being the logarithm with base ee and log\log as the logarithm with base 10.10. Thus, we can simplify the factors eln(34)e^{\ln (3^4)} and 10log(x)10^{\log(x)} using the inverse properties of logarithms. eln(34)10log(x)log5(125)=log(100)34xlog5(125)=log(100)\begin{aligned} \dfrac{e^{\ln (3^4)}\cdot 10^{\log(x)}}{\log_5\left( 125 \right)}&=\log(100) \\ \Updownarrow \\ \dfrac{3^4\cdot x}{\log_5\left( 125 \right)}&=\log(100) \end{aligned} To simplify the remaining two logarithms, we first need to rewrite their arguments. 34xlog5(125)=log(100)34xlog5(53)=log(102) \small \dfrac{3^4\cdot x}{\log_5\left( 125 \right)}=\log(100) \, \, \Leftrightarrow \, \, \dfrac{3^4\cdot x}{\log_5\left( 5^3 \right)}=\log\left( 10^2 \right) Again, we can use the the inverse properties of logarithms to simplify the equation. 34xlog5(53)=log(102)34x3=2 \dfrac{3^4\cdot x}{\log_5\left( 5^3 \right)}=\log\left( 10^2 \right) \, \, \Leftrightarrow \, \, \dfrac{3^4\cdot x}{3}=2 We now need to isolate xx on the left-hand side to find the solution.
34x3=2\dfrac{3^4\cdot x}{3}=2
33x=23^3\cdot x=2
x=233x=\dfrac{2}{3^3}
x=227x=\dfrac{2}{27}
We have solved the equation, and its solution is 227.\frac{2}{27}.


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