# Homework Help: Complex Logarithm and advanced Math Questions

1. Aug 21, 2011

### jspen30

Hello

I am currently studying Math B in high school and am having an extremely hard time with my current assessment. If anyone could please help me with the equations listed below it would be very much appreciated as my current Math teacher is on long service leave, and our substitute is practically teacher herself at the same time as teaching our class.

Equation 1:

Log 50 - 2 Log 5 - Log 2

Equation 2:

Log (x-3) = 1 + Log 4 - Log x

Equation 3:

2-4a + 2 x 2-2a - 8 = 0

Equation 4:

Without using calculator, find the value of A that makes x = e^12 / 1-2e^12 a solution of the equation:

ln x - ln (ax+1) = 12

I have attempted the first few on my own, how ever have not found the correct answer when checked against that of my calculator. They are listed below

Equation 1:

y=Log 50 - 2 Log 5 - Log 2
= 2 Log (50/5/2)
= 2 Log 5

Answer should be in Index form however I cannot make this into index form

I'm sorry for such a long post, and dont expect to be given the questions directly, however any help would be appreciated, even if it just pointing me in the right direction or explaining how to complete them. I'm very grateful for any help you may provide

Thankyou

Jake

Last edited: Aug 21, 2011
2. Aug 21, 2011

### Dick

Now what rules did you use to change Log 50 - 2 Log 5 - Log 2 into 2 Log 5? Can you spell it out?

3. Aug 21, 2011

### jspen30

i believe i used the law

Loga xy = loga x + logay

However I am not one hundred percent sure as my math teacher is currently away on stress leave (I doubt it was my class), and therefore I may be missing some formulae or equations to compete these questions

Jake

4. Aug 21, 2011

### Dick

What did you do with the '2' in front of '2 Log 5'? Use a*log(b)=log(b^a).

5. Aug 21, 2011

### jspen30

it was placed out the front eg

y=Log 50 - 2 Log 5 - Log 2
= 2Log (50/5/2)
= 2 Log 5

however this is not the final answer, as that is what im trying to find out, the final answer should be in index form

im trying to work out how i would do this equation to end up with an answer in index form, or if there is someway of going from the answer

2 Log 5 --- to index form

Thanks

6. Aug 21, 2011

### Dick

Here's an example. 1+2*3+4 is not equal to 2*(1+3+4). You can't pull the 2 on the second term out in front of the expression. That's an algebra mistake.

7. Aug 21, 2011

### jspen30

ok thankyou for telling me this mistake, How would I be able to find the following, showing full working and without using a calculator?

y = Log 50 - 2 Log 5 - Log 2
= 2 Log (50/5/2)
= 2 Log 5

Thanks again

Jake

8. Aug 21, 2011

### Dick

Use a*log(b)=log(b^a). So 2*log(5)=log(5^2), right? Now apply your other law.

9. Aug 21, 2011

### jspen30

ok thankyou

my current working out is:

Log 50 - Log 5^2 - Log 2
Log (50 / 5^2 / 2)
=1

Is this correct?

Jake

10. Aug 21, 2011

### Dick

You mean Log(1), right? That's not 1.

11. Aug 21, 2011

### jspen30

yes im very sorry

Log 50 - Log 5^2 - Log 2
Log (50 / 5^2 / 2)
= Log 1
= 0

is that correct working?

12. Aug 21, 2011

### Dick

Yes, it is. You don't usually want to write something like 50/25/2, though I know what you mean. Best to use some parentheses. (50/25)/2=1, 50/(25/2)=4. Not the same. The first one is the correct way to write it. Or 50/(25*2).

13. Aug 21, 2011

### jspen30

ok ill re-adjust my notation, thanks again for all the help, you have just taught me in about 5mins what my current substitute couldn't teach my class in 5 lessons. Do you have the knowledge to help me with any of the other equations listed above. If you dont have alot of spare time or have more important things to do I understand.

Thanks

Jake

14. Aug 21, 2011

### Dick

The most important thing I have to do right now is go to sleep. But I'll get you started. 'Log' with any indicated base often means log to the base e. If that's the case then take e to the power of both sides. Use e^(Log(a))=a.

15. Aug 21, 2011

### jspen30

ok thanks for all your help, and sorry for keeping you up (I live in Australia, and its only 2pm)

Cheers

Jake

16. Aug 21, 2011

### Dick

You aren't keeping me up. I'm keeping myself up. I'm sure somebody else can give you some help if I'm not around. Keep working on them and show your attempts.

17. Aug 22, 2011

### Redbelly98

Staff Emeritus
Last edited: Aug 22, 2011