# 6.5 Logarithmic properties  (Page 6/10)

 Page 6 / 10

Given a logarithm with the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}M,$ use the change-of-base formula to rewrite it as a quotient of logs with any positive base $\text{\hspace{0.17em}}n,$ where $\text{\hspace{0.17em}}n\ne 1.$

1. Determine the new base $\text{\hspace{0.17em}}n,$ remembering that the common log, $\text{\hspace{0.17em}}\mathrm{log}\left(x\right),$ has base 10, and the natural log, $\text{\hspace{0.17em}}\mathrm{ln}\left(x\right),$ has base $\text{\hspace{0.17em}}e.$
2. Rewrite the log as a quotient using the change-of-base formula
• The numerator of the quotient will be a logarithm with base $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ and argument $\text{\hspace{0.17em}}M.$
• The denominator of the quotient will be a logarithm with base $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ and argument $\text{\hspace{0.17em}}b.$

## Changing logarithmic expressions to expressions involving only natural logs

Change $\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\text{\hspace{0.17em}}$ to a quotient of natural logarithms.

Because we will be expressing $\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\text{\hspace{0.17em}}$ as a quotient of natural logarithms, the new base, $\text{\hspace{0.17em}}n=e.$

We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.

$\begin{array}{ll}{\mathrm{log}}_{b}M\hfill & =\frac{\mathrm{ln}M}{\mathrm{ln}b}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\hfill & =\frac{\mathrm{ln}3}{\mathrm{ln}5}\hfill \end{array}$

Change $\text{\hspace{0.17em}}{\mathrm{log}}_{0.5}8\text{\hspace{0.17em}}$ to a quotient of natural logarithms.

$\frac{\mathrm{ln}8}{\mathrm{ln}0.5}$

Can we change common logarithms to natural logarithms?

Yes. Remember that $\text{\hspace{0.17em}}\mathrm{log}9\text{\hspace{0.17em}}$ means $\text{\hspace{0.17em}}{\text{log}}_{\text{10}}\text{9}.$ So, $\text{\hspace{0.17em}}\mathrm{log}9=\frac{\mathrm{ln}9}{\mathrm{ln}10}.$

## Using the change-of-base formula with a calculator

Evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(10\right)\text{\hspace{0.17em}}$ using the change-of-base formula with a calculator.

According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base $\text{\hspace{0.17em}}e.$

Evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(100\right)\text{\hspace{0.17em}}$ using the change-of-base formula.

$\frac{\mathrm{ln}100}{\mathrm{ln}5}\approx \frac{4.6051}{1.6094}=2.861$

Access these online resources for additional instruction and practice with laws of logarithms.

## Key equations

 The Product Rule for Logarithms ${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$ The Quotient Rule for Logarithms ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$ The Power Rule for Logarithms ${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$ The Change-of-Base Formula

## Key concepts

• We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See [link] .
• We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See [link] .
• We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See [link] , [link] , and [link] .
• We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See [link] , [link] , and [link] .
• The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See [link] , [link] , [link] , and [link] .
• We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula. See [link] .
• The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ as the quotient of natural or common logs. That way a calculator can be used to evaluate. See [link] .

## Verbal

How does the power rule for logarithms help when solving logarithms with the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(\sqrt[n]{x}\right)?$

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left({x}^{\frac{1}{n}}\right)=\frac{1}{n}{\mathrm{log}}_{b}\left(x\right).$

What does the change-of-base formula do? Why is it useful when using a calculator?

## Algebraic

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

${\mathrm{log}}_{b}\left(7x\cdot 2y\right)$

${\mathrm{log}}_{b}\left(2\right)+{\mathrm{log}}_{b}\left(7\right)+{\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)$

$\mathrm{ln}\left(3ab\cdot 5c\right)$

${\mathrm{log}}_{b}\left(\frac{13}{17}\right)$

${\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)$

$\mathrm{ln}\left(\frac{1}{{4}^{k}}\right)$

$-k\mathrm{ln}\left(4\right)$

${\mathrm{log}}_{2}\left({y}^{x}\right)$

For the following exercises, condense to a single logarithm if possible.

$\mathrm{ln}\left(7\right)+\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)$

$\mathrm{ln}\left(7xy\right)$

${\mathrm{log}}_{3}\left(2\right)+{\mathrm{log}}_{3}\left(a\right)+{\mathrm{log}}_{3}\left(11\right)+{\mathrm{log}}_{3}\left(b\right)$

${\mathrm{log}}_{b}\left(28\right)-{\mathrm{log}}_{b}\left(7\right)$

${\mathrm{log}}_{b}\left(4\right)$

$\mathrm{ln}\left(a\right)-\mathrm{ln}\left(d\right)-\mathrm{ln}\left(c\right)$

$-{\mathrm{log}}_{b}\left(\frac{1}{7}\right)$

${\text{log}}_{b}\left(7\right)$

$\frac{1}{3}\mathrm{ln}\left(8\right)$

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

$\mathrm{log}\left(\frac{{x}^{15}{y}^{13}}{{z}^{19}}\right)$

$15\mathrm{log}\left(x\right)+13\mathrm{log}\left(y\right)-19\mathrm{log}\left(z\right)$

$\mathrm{ln}\left(\frac{{a}^{-2}}{{b}^{-4}{c}^{5}}\right)$

$\mathrm{log}\left(\sqrt{{x}^{3}{y}^{-4}}\right)$

$\frac{3}{2}\mathrm{log}\left(x\right)-2\mathrm{log}\left(y\right)$

$\mathrm{ln}\left(y\sqrt{\frac{y}{1-y}}\right)$

$\mathrm{log}\left({x}^{2}{y}^{3}\sqrt[3]{{x}^{2}{y}^{5}}\right)$

$\frac{8}{3}\mathrm{log}\left(x\right)+\frac{14}{3}\mathrm{log}\left(y\right)$

For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

$\mathrm{log}\left(2{x}^{4}\right)+\mathrm{log}\left(3{x}^{5}\right)$

$\mathrm{ln}\left(6{x}^{9}\right)-\mathrm{ln}\left(3{x}^{2}\right)$

$\mathrm{ln}\left(2{x}^{7}\right)$

$2\mathrm{log}\left(x\right)+3\mathrm{log}\left(x+1\right)$

$\mathrm{log}\left(x\right)-\frac{1}{2}\mathrm{log}\left(y\right)+3\mathrm{log}\left(z\right)$

$\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)$

$4{\mathrm{log}}_{7}\left(c\right)+\frac{{\mathrm{log}}_{7}\left(a\right)}{3}+\frac{{\mathrm{log}}_{7}\left(b\right)}{3}$

For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.

${\mathrm{log}}_{7}\left(15\right)\text{\hspace{0.17em}}$ to base $\text{\hspace{0.17em}}e$

${\mathrm{log}}_{7}\left(15\right)=\frac{\mathrm{ln}\left(15\right)}{\mathrm{ln}\left(7\right)}$

${\mathrm{log}}_{14}\left(55.875\right)\text{\hspace{0.17em}}$ to base $\text{\hspace{0.17em}}10$

For the following exercises, suppose $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(6\right)=a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(11\right)=b.\text{\hspace{0.17em}}$ Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ Show the steps for solving.

${\mathrm{log}}_{11}\left(5\right)$

${\mathrm{log}}_{11}\left(5\right)=\frac{{\mathrm{log}}_{5}\left(5\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{1}{b}$

${\mathrm{log}}_{6}\left(55\right)$

${\mathrm{log}}_{11}\left(\frac{6}{11}\right)$

${\mathrm{log}}_{11}\left(\frac{6}{11}\right)=\frac{{\mathrm{log}}_{5}\left(\frac{6}{11}\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{{\mathrm{log}}_{5}\left(6\right)-{\mathrm{log}}_{5}\left(11\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{a-b}{b}=\frac{a}{b}-1$

## Numeric

For the following exercises, use properties of logarithms to evaluate without using a calculator.

${\mathrm{log}}_{3}\left(\frac{1}{9}\right)-3{\mathrm{log}}_{3}\left(3\right)$

$6{\mathrm{log}}_{8}\left(2\right)+\frac{{\mathrm{log}}_{8}\left(64\right)}{3{\mathrm{log}}_{8}\left(4\right)}$

$3$

$2{\mathrm{log}}_{9}\left(3\right)-4{\mathrm{log}}_{9}\left(3\right)+{\mathrm{log}}_{9}\left(\frac{1}{729}\right)$

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

${\mathrm{log}}_{3}\left(22\right)$

$2.81359$

${\mathrm{log}}_{8}\left(65\right)$

${\mathrm{log}}_{6}\left(5.38\right)$

$0.93913$

${\mathrm{log}}_{4}\left(\frac{15}{2}\right)$

${\mathrm{log}}_{\frac{1}{2}}\left(4.7\right)$

$-2.23266$

## Extensions

Use the product rule for logarithms to find all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values such that $\text{\hspace{0.17em}}{\mathrm{log}}_{12}\left(2x+6\right)+{\mathrm{log}}_{12}\left(x+2\right)=2.\text{\hspace{0.17em}}$ Show the steps for solving.

Use the quotient rule for logarithms to find all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values such that $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)=1.\text{\hspace{0.17em}}$ Show the steps for solving.

$x=4;\text{\hspace{0.17em}}$ By the quotient rule: ${\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)={\mathrm{log}}_{6}\left(\frac{x+2}{x-3}\right)=1.$

Rewriting as an exponential equation and solving for $\text{\hspace{0.17em}}x:$

$\begin{array}{ll}{6}^{1}\hfill & =\frac{x+2}{x-3}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2}{x-3}-6\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2}{x-3}-\frac{6\left(x-3\right)}{\left(x-3\right)}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2-6x+18}{x-3}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x-4}{x-3}\hfill \\ \text{​}\text{\hspace{0.17em}}x\hfill & =4\hfill \end{array}$

Checking, we find that $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(4+2\right)-{\mathrm{log}}_{6}\left(4-3\right)={\mathrm{log}}_{6}\left(6\right)-{\mathrm{log}}_{6}\left(1\right)\text{\hspace{0.17em}}$ is defined, so $\text{\hspace{0.17em}}x=4.$

Can the power property of logarithms be derived from the power property of exponents using the equation $\text{\hspace{0.17em}}{b}^{x}=m?\text{\hspace{0.17em}}$ If not, explain why. If so, show the derivation.

Prove that $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(n\right)=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}\text{\hspace{0.17em}}$ for any positive integers $\text{\hspace{0.17em}}b>1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n>1.$

Let $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ be positive integers greater than $\text{\hspace{0.17em}}1.\text{\hspace{0.17em}}$ Then, by the change-of-base formula, $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(n\right)=\frac{{\mathrm{log}}_{n}\left(n\right)}{{\mathrm{log}}_{n}\left(b\right)}=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}.$

Does $\text{\hspace{0.17em}}{\mathrm{log}}_{81}\left(2401\right)={\mathrm{log}}_{3}\left(7\right)?\text{\hspace{0.17em}}$ Verify the claim algebraically.

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
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salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
hii
Uday
hi
salma
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Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar