# I know the very basic principles of logs

## Homework Statement

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'

6^(x+3)=56.34

## 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 :)

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Good. Now you have an equation of the form

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

Solve for x.

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

Solve for x.
So, the next line would be xlog6 +3log6 = 1.751

Yes or no?

BruceW
Homework Helper

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.

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

Yes or no?
Yes.

Mark44
Mentor

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
------
As already noted, the first line above has a typo.
So, the next line would be xlog6 +3log6 = 1.751

Yes or no?
Yes.
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.

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 :(

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

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 ?

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.

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

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

BruceW
Homework Helper

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.

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?

Mark44
Mentor

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?
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>.

cepheid
Staff Emeritus
Gold Member

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

You cannot put equals signs inside of expressions within parentheses. It makes no sense.
Sorry, not quite sure what you mean here?

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>.
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 ! :(

cepheid
Staff Emeritus
Gold Member

Sorry, not quite sure what you mean here?

$$(xlog75.2=3x-5)/log10.5$$
You have an equals sign inside the parentheses. It makes no sense.

BruceW
Homework Helper

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

BruceW
Homework Helper

CONFUSED ! :(
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.

BruceW
Homework Helper

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?
This bit is correct. Just think of the numbers as letters, you have something like: $ax = 3bx - 5b$, which is elementary algebra.

This bit is correct. Just think of the numbers as letters, you have something like: $ax = 3bx - 5b$, which is elementary algebra.
Ok... Thanks, I'll have another go now, My brain is fried haha!

BruceW
Homework Helper

No worries, I know the feeling

I really don't see what I can do here, nothing seems to look right! I can see on your equation it would be $$(x=3bx-5b)/a$$ but I just can't see how to apply that to my equation :(

BruceW
Homework Helper

Your equation is $xlog(75.2)=3xlog(10.5) - 5log(10.5)$ So you could also write it as $ax=bx+c$ if you wanted. You can choose how to define your constants.

You don't even need to write it as algebraic letters. You could just write down the actual numbers straight away, then find what x must be.