Using Properties of Logarithms

[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

 

[SammyS]

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]

What is the prime factorization of 250 ?

2, 5

 

[fresh_42]

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.

 

[fresh_42]

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

 

[SammyS]

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 ?

Added in Edit:

I see you have it now.

 

[fresh_42]

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?

 

[fresh_42]

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?

 

[SammyS]

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

 

[SammyS]

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?

 

[Mark44]

Okay thanks guys, so...

$\frac{1}{{5^{3}}\cdot2}$

The line above should be $\log_5 \frac{1}{{5^{3}}\cdot2}$

= $log_{5}5^{3}-log_{5}5^{2}$

No. The thing you're taking the log of is a quotient that also includes a product.

= $3log_{5}5 - 2log_{5}5$

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

You're using 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]

Okay, guys. This is as far as my knowledge takes me. 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.

 

[Mark44]

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.

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.

 

[Mark44]

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.

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]

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]

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!