# 4.5 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.

can you not take the square root of a negative number
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
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Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas