# Using Properties of Logarithms

• FritoTaco
In summary, the conversation was about the importance of effective communication in a workplace setting. The participants discussed the benefits of clear and concise communication, as well as the negative effects of miscommunication. They also touched on strategies for improving communication, such as active listening and being mindful of nonverbal cues. Overall, the conversation emphasized the crucial role that communication plays in creating a positive and productive work environment.
FritoTaco

## Homework Statement

[/B]
Question: Use the properties of logarithms to rewrite and simplify the logarithmic expression.

1. $$log_{5}\dfrac{1}{250}$$

## Homework Equations

Product Property: $$log_{a}(uv) = log_{a}u + log_{a}v$$
Quotient Property: $$log_{a}\dfrac{u}{v} = log_{a}u - log_{a}v$$
Power Property: $$log_{a}u^{k} = k\cdot log_{a}u$$

## The Attempt at a Solution

$$log_{5}1 - log_{5}250$$

I know when it's dividing, you split it up by subtracting. (Still not confident in the first step) Do I find out 5 to what power gives it 250? Even if I did that, it won't match 250. Pretty sure I'm missing something I'm unaware of.

Answer: $$-3 - log_{5}2$$

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FritoTaco said:

## Homework Statement

[/B]
Question: Use the properties of logarithms to rewrite and simplify the logarithmic expression.

1. $$log_{5}\dfrac{1}{250}$$

## Homework Equations

Product Property: $$log_{a}(uv) = log_{a}u + log_{a}v$$
Quotient Property: $$log_{a}\dfrac{u}{v} = log_{a}u - log_{a}v$$
Power Property: $$log_{a}u^{k} = k\cdot log_{a}u$$

## The Attempt at a Solution

$$log_{5}1 - log_{5}250$$

I know when it's dividing, you split it up by subtracting. (Still not confident in the first step) Do I find out 5 to what power gives it 250? Even if I did that, it won't match 250. Pretty sure I'm missing something I'm unaware of.

Answer: $$-3 - log_{5}2$$
What is the prime factorization of 250 ?

FritoTaco
SammyS said:
What is the prime factorization of 250 ?

2, 5

Consider the integer factorization. And how else can you write ##\frac{1}{250}##?

FritoTaco
Maybe this: $$5^{3} \cdot 2$$

I don't know how to flip it so it's 1/250. That's why I separated them.

Better. And now ##\frac{1}{5^3\,\cdot\,2}## and your rules can be applied.

FritoTaco
FritoTaco said:
2, 5
No. Those are the prime factors, but 2⋅5 = 10, not 250.

Do a factor tree.

How many factors of 2 and how many factors of 5 ?

I see you have it now.

FritoTaco
FritoTaco said:
Maybe this: $$5^{3} \cdot 2$$

I don't know how to flip it so it's 1/250. That's why I separated them.
Not that important. You can use rule two as well. Just know what ##\log_5 1## and ##\log_5 5## is.

FritoTaco
Okay thanks guys, so...

$$\dfrac{1}{{5^{3}}\cdot2}$$
= $$log_{5}5^{3}-log_{5}5^{2}$$
= $$3log_{5}5 - 2log_{5}5$$

I'm closer, but still stuck on getting the correct answer, is this good so far?

FritoTaco said:
Okay thanks guys, so...

$$\dfrac{1}{{5^{3}}\cdot2}$$
= $$log_{5}5^{3}-log_{5}5^{2}$$
That's not what the quotient rule says, and what you did right the first time.
= $$3log_{5}5 - 2log_{5}5$$

I'm closer, but still stuck on getting the correct answer, is this good so far?
Apply them one at a time: quotient rule - product rule - special cases. And don't forget parentheses when in doubt.

FritoTaco
Ok, so I see an error. I have a 5 base 5 in the second half of step 2. I think it should be $$log_{5}5^{3}-log_{5}2$$, then $$3log_{5}5^{}-log_{5}2$$. Okay, but how do I get rid of the 5 and have a negative sign in front of the 3?

FritoTaco said:
Okay thanks guys, so...
$$\dfrac{1}{{5^{3}}\cdot2}$$ $$=log_{5}5^{3}-log_{5}5^{2}$$...

I'm closer, but still stuck on getting the correct answer, is this good so far?
Your mistakes are in the quoted step.

You used the quotient rule correctly in the OP, when you obtained
## \log_{5}1 - \log_{5}250 ##​
Now use the factorization of 250: 2⋅53 to deconstruct log5(250).

Notice:
2 is not an exponent. It's a coefficient.​
.

FritoTaco
Okay, so I have this so far: $$log_{5}1-log_{5}250$$ $$log_{5}1-log_{5}(5^{3}\cdot2)$$ $$log_{5}5^{3}-log_{5}2$$ $$3log_{5}5-log_{5}2$$

FritoTaco said:
Okay, so I have this so far: $$log_{5}1-log_{5}250$$ $$log_{5}1-log_{5}(5^{3}\cdot2)$$ $$log_{5}5^{3}-log_{5}2$$ $$3log_{5}5-log_{5}2$$
Look at ##\ \log_5(250) = ... \ ## by itself. Then consider that it all comes from the denominator.

FritoTaco
Does that mean if it comes from denominator the whole solution becomes negative?

FritoTaco said:
Okay thanks guys, so...

##\frac{1}{{5^{3}}\cdot2}##
The line above should be ##\log_5 \frac{1}{{5^{3}}\cdot2}##
FritoTaco said:
= ##log_{5}5^{3}-log_{5}5^{2}##
No. The thing you're taking the log of is a quotient that also includes a product.
FritoTaco said:
= ##3log_{5}5 - 2log_{5}5##

I'm closer, but still stuck on getting the correct answer, is this good so far?
You're using [tex] tags, which causes the rendered expressions to be centered on their own line. For these equations, $(inline LaTeX) would look better, IMO. It's simpler to surround the material with  (standalone) or ## (inline) tags. FritoTaco and jim mcnamara Okay, guys. This is as far as my knowledge takes me. [itex]log_{5}\dfrac{1}{250}$
$log_{5}1-log_{5}(5^{3}\cdot2)$
$log_{5}1-log_{5}5^{3}+log_{5}2$
$log_{5}1-3log_{5}5+log_{5}2$ (log 5 base 5 cancels out)
$-3+log_{5}2$
So which step did I go wrong? Also, thanks for the information, I was wondering how to format it.

FritoTaco said:
Okay, guys. This is as far as my knowledge takes me. $log_{5}\dfrac{1}{250}$
$log_{5}1-log_{5}(5^{3}\cdot2)$
Above is fine, but why not simplify it instead of dragging along ##\log_5(1)##? Do you know what this equals?
The work below contains an error. The fact that you're dragging along ##\log_5(1)## probably contributed to this error.
FritoTaco said:
$log_{5}1-log_{5}5^{3}+log_{5}2$
$log_{5}1-3log_{5}5+log_{5}2$ (log 5 base 5 cancels out)
$-3+log_{5}2$
So which step did I go wrong? Also, thanks for the information, I was wondering how to format it.

FritoTaco
Doesn't $log_{5}1$ equal zero? I ignored it when solving the rest of the equation.

FritoTaco said:
Doesn't $log_{5}1$ equal zero? I ignored it when solving the rest of the equation.
Yes, that's the right value.

Here's the work you showed in post#17.
FritoTaco said:
$log_{5}1-log_{5}(5^{3}\cdot2)$
$log_{5}1-log_{5}5^{3}+log_{5}2$
$log_{5}1-3log_{5}5+log_{5}2$ (log 5 base 5
Dragging ##\log_5(1)## along in each of these steps probably caused the error I mentioned.

BTW you are not "solving the equation." You are evaluating an expression; namely ##\log_5(\frac 1 {250})##. Solving an equation and evaluating or simplifying an expression are two completely different concepts.

Here's a better start to this problem:
##\log_5(\frac 1 {250})##
##= \log_5(1) - \log_5(250)##
## = -\log_5(250)##
## = \dots ##

FritoTaco
Yes! It has finally arrived!:

$log_{5}(\dfrac{1}{250})$
= $log_{5}(1)-log_{5}(250)$
= $-log_{5}(250)$
= $-log_{5}(5^{3}\cdot2)$
= $-log_{5}5^3-log_{5}2$
= $-3-log_{5}2$

Thank you for your help people!

SammyS
FritoTaco said:
Yes! It has finally arrived!:

$log_{5}(\dfrac{1}{250})$
= $log_{5}(1)-log_{5}(250)$
= $-log_{5}(250)$
= $-log_{5}(5^{3}\cdot2)$
= $-log_{5}5^3-log_{5}2$
= $-3-log_{5}2$

Thank you for your help people!
Very good !

I think that one thing @fresh_42 was suggesting way back in post #4 or thereabouts was to write 1/(250) as (250)-1 .

Then you start with ##\ \displaystyle \log_5\left(\frac 1 {250} \right) = \log_5 (250^{-1})=(-1) \log_5 (250)= \ \dots##

FritoTaco
SammyS said:
Very good !

I think that one thing @fresh_42 was suggesting way back in post #4 or thereabouts was to write 1/(250) as (250)-1 .

Then you start with ##\ \displaystyle \log_5\left(\frac 1 {250} \right) = \log_5 (250^{-1})=(-1) \log_5 (250)= \ \dots##
Oh, so that's another way to start. That makes sense. Time to practice more problems!

## 1. What are the basic properties of logarithms?

The basic properties of logarithms are:
1. The power property: logb(xy) = y * logb(x)
2. The product property: logb(x * y) = logb(x) + logb(y)
3. The quotient property: logb(x / y) = logb(x) - logb(y)
4. The change of base property: logb(x) = loga(x) / loga(b)
5. The inverse property: logb(bx) = x

## 2. How do you simplify logarithmic expressions using properties?

To simplify logarithmic expressions using properties, you can use the basic properties mentioned above. Start by applying the power property to any exponents in the expression. Then, use the product and quotient properties to combine any logarithms that have the same base. Finally, use the change of base property to convert the logarithm to a different base if needed. Remember to check for any restrictions on the domain of the logarithm before simplifying.

## 3. Can properties of logarithms be used to solve equations?

Yes, properties of logarithms can be used to solve equations involving logarithms. To solve an equation with a logarithm, use the properties to simplify the expression and isolate the logarithm on one side of the equation. Then, use the inverse property to rewrite the equation in exponential form. Finally, solve for the variable using algebraic techniques.

## 4. Are there any special properties of logarithms?

Yes, there are a few special properties of logarithms that are frequently used in solving logarithmic equations. These include:
1. The zero property: logb(1) = 0
2. The identity property: logb(b) = 1
3. The negative property: logb(x) is undefined for x ≤ 0

## 5. How can properties of logarithms be applied in real-life situations?

Properties of logarithms can be applied in various real-life situations, such as:
1. Calculating the pH of a solution in chemistry
2. Measuring the loudness of sound in decibels
3. Analyzing the growth and decay of populations in biology
4. Calculating the intensity of earthquakes on the Richter scale
5. Estimating the amount of time needed for an investment to double in value in finance
6. Determining the amount of time needed for a medication to reach a therapeutic level in pharmacology
7. Solving exponential growth and decay problems in economics and business

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