# 2 Quick log problems

1. Feb 8, 2010

### lexpar

I'm new here. You already knew that. Thank you, you, for clicking on my thread in any case. So, I've got these two log problems. My math teacher gave them out as a challenge. The class is relatively new to logs, and I'm having trouble working with these longer questions (longer than "simplify log5+log3"). I think my main issue is with the sub scripts. I don't really know what to do with them. For the first question, where it a normal algebra question, I'd have it no problem. Those logs get in the way though. We've been given the answers, but need to prove them. Any help would be appreciated.

PS: the problems aren't for marks, they're more of a challenge thing before we do our test.

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

1: log25+log2(x-3)=4

3. The attempt at a solution

log25+log2(x-3)=4
log25(x-3)=4
log2(5x-15)=4
???
???
x= 31/5

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

log3(x+4)=log77-log3(x+2)

3. The attempt at a solution

log3(x+4)=log77-log3(x+2)
???
???
???
x= -9
x=-5

2. Feb 8, 2010

### Staff: Mentor

In your first problem, use the fact that log2a = b <===> a = 2b.
In your second problem, transpose the log expression on the right side over to the left side, and use the fact that log77 = 1. Then you'll have a problem very similar to your first problem.

3. Feb 8, 2010

### Staff: Mentor

Try to just apply definition of log.

If

$$log_ab = c$$

then

$$b = a^c$$

Edit: Mark beat me.

4. Feb 8, 2010

### tiny-tim

Welcome to PF!

Hi lexpar! Welcome to PF!
If this was ln(5x-15) = 4, you'd just do eboth sides, wouldn't you?

ok, so do 2both sides.
Hint: log77 = … ?

5. Feb 8, 2010

### lexpar

Thanks all three of you for the ultra speedy reply :D

Figured out the first:

log25+log2(x-3)=4
log25(x-3)=4
log2(5x-15)=4
log25+log2x-log215=log216
log25+log2x=log231
5x=31
x=31/5!!!! :D

As for the second:

log3(x+4)=log77-log3(x+2)
log3((x+4)(x+2))= 1 (or maybe log33?)

Where to go from here?

6. Feb 8, 2010

### tiny-tim

Same method as for the first one!

7. Feb 8, 2010

### Staff: Mentor

Unfortunately - twice wrong.

log(a-b) is NOT log(a) - log(b).

8. Feb 8, 2010

### lexpar

In that case, what's the correct process? 31/5 is the correct answer, or so we were told.

EDIT: Put a lot of thought into it. Read your earlier posts again:

log2(5x-15)=4
logab=c then b=ac
therefore
5x-15=24
x=31/5

Better? :D

Last edited: Feb 8, 2010
9. Feb 8, 2010

### tiny-tim

Just write log2(5x-15) = log216

10. Feb 8, 2010

### Staff: Mentor

31/5 is a correct answer, but you got it by accident.

Try 5x-15 = ... (just apply the definition, nothing more).

11. Feb 8, 2010

### lexpar

Thanks, I got it :D Check out my last post, I edited it just for you <3
*ahem*

So still having a bit of trouble with the second question. I get to:

log3((x+4)(x+2)=1
log3(x2+6x+8)=1
apply the definition=
x2+6x+8 = 3

But now I've got 2 x's! Where have I gone wrong??

12. Feb 8, 2010

### Staff: Mentor

You listed two x's as correct answers. This is quadratic function. So everything seems to fit.

Apart from the answers, seems to me like -9 is wrong. But I can be wrong, it is 1 a.m. here.

13. Feb 8, 2010

### Staff: Mentor

So far, so good. Add -3 to both sides, and solve the quadratic equation. It's an easy one so you can factor it.

Be sure to check both of your answers in the original equation, because one of them is extraneous.

14. Feb 8, 2010

### lexpar

Thank you all for the responses. You guys really run a quality forum here :D

Can I tempt you guys with one more problem? This one wasn't assigned, I'm just curious. I'm looking for a step by step walkthru here if you guys can give it, since this one flies right over my head.

Question:

logc20c/(logc50-logc5)

My progress:

logc20c/(logc50-logc5)
logc20c/(logc10)
...

PS: sorry for the late response, was out playing pool with my dad. It's a very important school night activity :p

Last edited: Feb 8, 2010
15. Feb 8, 2010

### Staff: Mentor

This is fine, but put an = between the two expressions. The numerator could be written as logc(20) + 1, but I don't see that as an improvement.

16. Feb 9, 2010

### tiny-tim

Hi lexpar!

(just got up :zzz: …)
Always remember that logca = ln(a)/ln(c) = log(a)/log(c) = logba/logbc for any b.

In this case, logc20c/(logc10) = … ?

17. Feb 9, 2010

### lexpar

...

I'm really not catching on to that one :C

18. Feb 9, 2010

### tiny-tim

Write each of the top and the bottom of logc20c/(logc10) as fractions.