# I know the very basic principles of logs

1. Dec 3, 2011

### lloydowen

1. The problem statement, all variables and given/known data
I'm new to logs. I know the very basic principles of logs, unfortunately my notes are not helping my on this occasion. What I have so far is Solve the value 'x'

2. Relevant equations

6^(x+3)=56.34

3. The attempt at a solution
Here's what I have done, and I'm not sure where to go from here so could someone explain?

log6^(x+3)=log5.34
(x+3)log6=log56.34
(x+3)log6=1.751
------

Not sure if I've done it correctly, and I'm not sure where to go from next :/

If you have seen me asking a lot of questions tonight, I have an exam next week and I'm just getting ready for it :)

2. Dec 3, 2011

### micromass

Staff Emeritus
Re: Logs/Logarithm

Good. Now you have an equation of the form

$$(x+3)b=c$$

Solve for x.

3. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

So, the next line would be xlog6 +3log6 = 1.751

Yes or no?

4. Dec 3, 2011

### BruceW

Re: Logs/Logarithm

I'm guessing on your first line, you meant log56.34 (since that's what you put on the next line, which is correct).

You have got it right so far. From here, its simple rearrangement to get x.

Edit: Sorry, I started writing this before micromass posted, so I am a little behind here.

yes, that would be the next line (although you could instead have divided both sides by b), but you get the right answer either way.

5. Dec 3, 2011

### micromass

Staff Emeritus
Re: Logs/Logarithm

Yes.

6. Dec 3, 2011

### Staff: Mentor

Re: Logs/Logarithm

As already noted, the first line above has a typo.
As an alternative, starting from (x + 3)log(6) = log(56.34),
you could divide both sides by log(6) to get x + 3 = log(56.34)/log(6), and you can solve for x from there.

Note that put back in log(56.34) instead of using your value of 1.751. You'll get a more precise result if you DON'T put in rounded intermediate values. If you use the full values that your calculator gives you, you'll get answers that don't have built-in inaccuracies.

7. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

So, on that last line I done... xlog6 + 3log6

Do I (on a calculator) type log6= then Ans*3... and then add it to log6 ? I'm confused now :(

8. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

Any help guys? I'm real stuck ! Can't move on from here...

9. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

So I couldn't do multiplication method so I went for division.. Here's what I got.

(x+3=1.751)/log6

Uhh doesn't look right but is it ?

10. Dec 3, 2011

### micromass

Staff Emeritus
Re: Logs/Logarithm

I really don't know what is bothering you so much. Can you solve

$$2(x+3)=5$$

??

The equation you list is the same thing but with different numbers.

11. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

So I think I have a final answer, I kept simplifying down. after that previous line i got (x=1.751)/log6 (-3)

Which then gave me a final answer of -0.75 to 2dp

12. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

New Question!!!

So I got another Question because I didn't full click with that last one..

75.2^x = 10.5 ^(3x-5)

What I got so far...

log75.2^x=log10.5^(3x-5)
xlog75.2 = (3x-5)log10.5

Trying the next line... Suppose its the same as last time but the other way around?

Edit:
$$(xlog75.2=3x-5)/log10.5$$

13. Dec 3, 2011

### BruceW

Re: Logs/Logarithm

You got the last question right, with -0.75.

For the new question, you have got the use of logarithm correct. Now it is just a case of rearranging.

14. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

The new question I re-wrote the equation like so, $$xlog75.2 = 3xlog10.5-5log10.5$$ Is that wrong and/or a better method than the last way?

15. Dec 3, 2011

### Staff: Mentor

Re: Logs/Logarithm

You're not done yet. Add -3xlog(10.5) to both sides. Then on the left side, you'll have two terms with a factor of x. Factor the two terms into x(...other stuff) = <stuff on the right side>.

16. Dec 3, 2011

### cepheid

Staff Emeritus
Re: Logs/Logarithm

You cannot put equals signs inside of expressions within parentheses. It makes no sense.

17. Dec 3, 2011

### lloydowen

Re: Logs/Logarithm

Sorry, not quite sure what you mean here?

Thanks for your help Mark, but I'm still unsure now :( Would it look like this..
$$(-3xlog(10.5)xlog75.2=-3xlog(10.5) 3x-5) / log10.5$$

Or $$((-3xlog(10.5)xlog75.2=(3xlog(10.5))3xlog10.5-5log10.5)$$ CONFUSED ! :(

18. Dec 3, 2011

### cepheid

Staff Emeritus
Re: Logs/Logarithm

You have an equals sign inside the parentheses. It makes no sense.

19. Dec 3, 2011

### BruceW

Re: Logs/Logarithm

haha, yeah, I think he means 'divide both sides by log10.5'

Edit: I don't know if that is allowable notation in maths, but you can see what he means by it.

20. Dec 3, 2011

### BruceW

Re: Logs/Logarithm

Think it through yourself, you've done the hard bit already, you just needed to rearrange to get x as the subject of the formula.