# Thread: Some simple logarithm questions

1. ## Some simple logarithm questions

the book simply says find the exact value of each expression. without a calculator. using properties of logarithms.( which it lists, but I can't see where they apply)
1. log6^18-log6^3
for this I just divided 18 by 3 getting 6. log6^6=1? is that right?
2.log3^8 * log8^9
for this i tried change of base. I did
(logb^8/logb^3) * (logb^9/logb^8)
3. e^(log(e^2)^16)
I'm completely lost on this one. Supposedly these are all really easy to do without a calculator. i just don't know how to do them. Any help would be greatly appreciated.

2. Originally Posted by bilbobaggins
the book simply says find the exact value of each expression. without a calculator. using properties of logarithms.( which it lists, but I can't see where they apply)
1. log6^18-log6^3
for this I just divided 18 by 3 getting 6. log6^6=1? is that right?
2.log3^8 * log8^9
for this i tried change of base. I did
(logb^8/logb^3) * (logb^9/logb^8)
3. e^(log(e^2)^16)
I'm completely lost on this one. Supposedly these are all really easy to do without a calculator. i just don't know how to do them. Any help would be greatly appreciated.
Are these logs to base ten?

If the one problem is this $\log_{6}6=y$

then by the inverse property we have $6^y=6\Longleftrightarrow{y=1}$

The logarithm of x to the base b is written logb(x) or, if the base is implicit, as log(x). So, for a number x, a base b and an exponent y,

An

3. Originally Posted by VonNemo19
Are these logs to base ten?

If the one problem is this $\log_{6}6=y$

then by the inverse property we have 6^y=6\Longleftrightarrow{y=1}
No they're not base 10. I'm sorry, the bases are supposed to be the numbers immediately after the log. I just didn't know how to write it on a computer.

4. Originally Posted by bilbobaggins
the book simply says find the exact value of each expression. without a calculator. using properties of logarithms.( which it lists, but I can't see where they apply)
1. log6^18-log6^3
for this I just divided 18 by 3 getting 6. log6^6=1? is that right?
2.log3^8 * log8^9
for this i tried change of base. I did
(logb^8/logb^3) * (logb^9/logb^8)
3. e^(log(e^2)^16)
I'm completely lost on this one. Supposedly these are all really easy to do without a calculator. i just don't know how to do them. Any help would be greatly appreciated.
1. Correct.

2. Change of base is the right way to go here but you pick a given value of b: usually base e or base 10:

$log_3{8} = \frac{ln8}{ln{3}} = \frac{3ln2}{ln3}$

$log_8{9} = \frac{ln{9}}{ln{8}} = \frac{2ln{3}}{3ln2}$

$\frac{3ln2}{ln{3}} \times \frac{2ln{3}}{3ln2} = 2$

3. Do you mean?

$e^{log_{e^2}16}$ -> the base of the log is e^2

it's an unusual base so I'd use the change of base as above ignoring the base of the exponent for now (the e at the start). I am choosing base e because then it will cancel at the end.

$log_{e^2}16$

$\frac{ln{16}}{ln{e^2}} = \frac{4ln(2)}{2ln(e)} = {2ln(2)} = ln(4)$

Bringing in our exponent we get $e^{ln(4)} = 4$

5. Properties:
Logarithmic identity

6. Originally Posted by e^(i*pi)
1. Correct.

2. Change of base is the right way to go here but you pick a given value of b: usually base e or base 10:

$log_3{8} = \frac{ln8}{ln{3}} = \frac{3ln2}{ln3}$

$log_8{9} = \frac{ln{9}}{ln{8}} = \frac{2ln{3}}{3ln2}$

$\frac{3ln2}{ln{3}} \times \frac{2ln{3}}{3ln2} = 2$

3. Do you mean?

$e^{log_{e^2}16}$ -> the base of the log is e^2

it's an unusual base so I'd use the change of base as above ignoring the base of the exponent for now (the e at the start). I am choosing base e because then it will cancel at the end.

$log_{e^2}16$

$\frac{ln{16}}{ln{e^2}} = \frac{4ln(2)}{2ln(e)} = {2ln(2)} = ln(4)$

Bringing in our exponent we get $e^{ln(4)} = 4$
So does an ln cancel out an e?

For example in these two.

lne^-4 and e^ln8
By the way, thank you so much for the help.

7. Originally Posted by bilbobaggins
So does an ln cancel out an e?

For example in these two.

lne^-4 and e^ln8
By the way, thank you so much for the help.
Yes, e and ln are inverse operators, like addition and subtraction etc...

$ln(x)$ is just a different way of saying $log_e(x)$

$ln(e^x) = e^{ln(x)} = x$

So $ln(e^{-4}) = -4$and $e^{ln(8)} = 8$

8. I have another one that's really confusing.

21log3^cuberootx +log3^(9x^2)-log3^9

I'm supposed to get it to a single logarithm. I get it down to log3^(x^20). Is that right? Or is it log3^(9x^20/9)?

9. Originally Posted by bilbobaggins
I have another one that's really confusing.

21log3^cuberootx +log3^(9x^2)-log3^9

I'm supposed to get it to a single logarithm. I get it down to log3^(x^20). Is that right? Or is it log3^(9x^20/9)?
$21log_3(x^{\frac{1}{3}}) = log_3(x^{\frac{21}{3}}) + log_3(9x^2) - log_3(9)$

$= log_3(7\times 9x^2 \div 9) = log_3(7x^2)$

That will need to be simplified to give a final answer

10. Originally Posted by e^(i*pi)
$21log_3(x^{\frac{1}{3}}) = log_3(x^{\frac{21}{3}}) + log_3(9x^2) - log_3(9)$

$= log_3(7\times 9x^2 \div 9) = log_3(7x^2)$

That will need to be simplified to give a final answer
I'm a little lost on the second line. So the x^21/3 becomes just seven? Because of 21/3?

11. Originally Posted by bilbobaggins
I'm a little lost on the second line. So the x^21/3 becomes just seven? Because of 21/3?
Damn, that's a typo, It should be x^7 rather than 7. The exponent of 7 does come from the 21/3

$
= log_3(x^7 \times 9x^2 \div 9) = log_3(x^9)
$

You multiply because of the rule that says log(a)+log(b) = log(ab). The 9/9 will cancel inside the log term

12. Originally Posted by e^(i*pi)
Damn, that's a typo, It should be x^7 rather than 7. The exponent of 7 does come from the 21/3

$
= log_3(x^7 \times 9x^2 \div 9) = log_3(x^9)
$

You multiply because of the rule that says log(a)+log(b) = log(ab). The 9/9 will cancel inside the log term
Alright, thank you very much for the help. It's a lot clearer now. I wish you were my math instructor!