# Using Properties of Logarithms

1. Oct 13, 2016

### FritoTaco

1. The problem statement, all variables and given/known data

Question: Use the properties of logarithms to rewrite and simplify the logarithmic expression.

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

2. Relevant 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: $$log_{a}u^{k} = k\cdot log_{a}u$$

3. 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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2. Oct 13, 2016

### SammyS

Staff Emeritus
What is the prime factorization of 250 ?

3. Oct 13, 2016

### FritoTaco

2, 5

4. Oct 13, 2016

### Staff: Mentor

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

5. Oct 13, 2016

### 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.

6. Oct 13, 2016

### Staff: Mentor

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

7. Oct 13, 2016

### SammyS

Staff Emeritus
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.

8. Oct 13, 2016

### Staff: Mentor

Not that important. You can use rule two as well. Just know what $\log_5 1$ and $\log_5 5$ is.

9. Oct 13, 2016

### 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?

10. Oct 13, 2016

### Staff: Mentor

That's not what the quotient rule says, and what you did right the first time.
Apply them one at a time: quotient rule - product rule - special cases. And don't forget parentheses when in doubt.

11. Oct 14, 2016

### 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?

12. Oct 14, 2016

### SammyS

Staff Emeritus
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.​
.

13. Oct 14, 2016

### FritoTaco

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

14. Oct 14, 2016

### SammyS

Staff Emeritus
Look at $\ \log_5(250) = ... \$ by itself. Then consider that it all comes from the denominator.

15. Oct 14, 2016

### FritoTaco

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

16. Oct 14, 2016

### Staff: Mentor

The line above should be $\log_5 \frac{1}{{5^{3}}\cdot2}$
No. The thing you're taking the log of is a quotient that also includes a product.