# 4.5 Logarithmic properties  (Page 5/10)

 Page 5 / 10

## Condensing complex logarithmic expressions

Condense $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x-1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right).$

We apply the power rule first:

${\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x-1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)={\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x-1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)$

Next we apply the product rule to the sum:

${\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x-1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\left({x}^{2}\sqrt{x-1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)$

Finally, we apply the quotient rule to the difference:

${\mathrm{log}}_{2}\left({x}^{2}\sqrt{x-1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\frac{{x}^{2}\sqrt{x-1}}{{\left(x+3\right)}^{6}}$

## Rewriting as a single logarithm

Rewrite $\text{\hspace{0.17em}}2\mathrm{log}x-4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)\text{\hspace{0.17em}}$ as a single logarithm.

We apply the power rule first:

$\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}{\left(x+5\right)}^{4}+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)$

Next we rearrange and apply the product rule to the sum:

$\mathrm{log}\left({x}^{2}\right)-\mathrm{log}{\left(x+5\right)}^{4}+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)$
$=\mathrm{log}\left({x}^{2}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)-\mathrm{log}{\left(x+5\right)}^{4}$
$=\mathrm{log}\left({x}^{2}{\left(3x+5\right)}^{{x}^{-1}}\right)-\mathrm{log}{\left(x+5\right)}^{4}$

Finally, we apply the quotient rule to the difference:

$=\mathrm{log}\left({x}^{2}{\left(3x+5\right)}^{{x}^{-1}}\right)-{\mathrm{log}\left(x+5\right)}^{4}=\mathrm{log}\frac{{x}^{2}{\left(3x+5\right)}^{{x}^{-1}}}{{\left(x+5\right)}^{4}}$

Rewrite $\text{\hspace{0.17em}}\mathrm{log}\left(5\right)+0.5\mathrm{log}\left(x\right)-\mathrm{log}\left(7x-1\right)+3\mathrm{log}\left(x-1\right)\text{\hspace{0.17em}}$ as a single logarithm.

$\mathrm{log}\left(\frac{5{\left(x-1\right)}^{3}\sqrt{x}}{\left(7x-1\right)}\right)$

Condense $\text{\hspace{0.17em}}4\left(3\mathrm{log}\left(x\right)+\mathrm{log}\left(x+5\right)-\mathrm{log}\left(2x+3\right)\right).$

$\mathrm{log}\frac{{x}^{12}{\left(x+5\right)}^{4}}{{\left(2x+3\right)}^{4}};\text{\hspace{0.17em}}$ this answer could also be written $\text{\hspace{0.17em}}\mathrm{log}{\left(\frac{{x}^{3}\left(x+5\right)}{\left(2x+3\right)}\right)}^{4}.$

## Applying of the laws of logs

Recall that, in chemistry, $\text{\hspace{0.17em}}\text{pH}=-\mathrm{log}\left[{H}^{+}\right].\text{\hspace{0.17em}}$ If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

Suppose $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is the original concentration of hydrogen ions, and $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is the original pH of the liquid. Then $\text{\hspace{0.17em}}P=–\mathrm{log}\left(C\right).\text{\hspace{0.17em}}$ If the concentration is doubled, the new concentration is $\text{\hspace{0.17em}}2C.$ Then the pH of the new liquid is

$\text{pH}=-\mathrm{log}\left(2C\right)$

Using the product rule of logs

$\text{pH}=-\mathrm{log}\left(2C\right)=-\left(\mathrm{log}\left(2\right)+\mathrm{log}\left(C\right)\right)=-\mathrm{log}\left(2\right)-\mathrm{log}\left(C\right)$

Since $\text{\hspace{0.17em}}P=–\mathrm{log}\left(C\right),$ the new pH is

$\text{pH}=P-\mathrm{log}\left(2\right)\approx P-0.301$

When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.

How does the pH change when the concentration of positive hydrogen ions is decreased by half?

The pH increases by about 0.301.

## Using the change-of-base formula for logarithms

Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or $\text{\hspace{0.17em}}e,$ we use the change-of-base formula    to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.

To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms    .

Given any positive real numbers $\text{\hspace{0.17em}}M,b,$ and $\text{\hspace{0.17em}}n,$ where and $\text{\hspace{0.17em}}b\ne 1,$ we show

$\text{\hspace{0.17em}}{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}$

Let $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}M.\text{\hspace{0.17em}}$ By taking the log base $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ of both sides of the equation, we arrive at an exponential form, namely $\text{\hspace{0.17em}}{b}^{y}=M.\text{\hspace{0.17em}}$ It follows that

For example, to evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{5}36\text{\hspace{0.17em}}$ using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.

## The change-of-base formula

The change-of-base formula    can be used to evaluate a logarithm with any base.

For any positive real numbers $\text{\hspace{0.17em}}M,b,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n,$ where and $\text{\hspace{0.17em}}b\ne 1,$

${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}.$

It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.

${\mathrm{log}}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b}$

and

${\mathrm{log}}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}$

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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
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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."
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Thomas
Darius
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Thomas
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Darius
I've been struggling so much through all of this. my final is in four weeks 😭
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So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
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Nare
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The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
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Am
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Am
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Case of Equilateral Hyperbola
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f(x)=4x+2, find f(3)
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f(3)=4(3)+2 f(3)=14
lamoussa
14
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Explain why log a x is not defined for a < 0
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Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
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Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
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Gilbert