Solve Simple Logarithms: log91/27

[matadorqk]

Thanks evrn!

Homework Statement

Express as a single logarithm 2lnx + ln (x-1) - ln (x-2)

Homework Equations

None used (Basic Algebra)

The Attempt at a Solution

So, let's substitute ln for a, to make it simpler
2ax +a (x-1) - a (x-2)
2ax +ax - a - ax +2a
2ax - a +2a
2ax+a
2lnx+ln
ln (2x+1)

I think that's correct, I'd appreciate some re-check just in case though.

That's the last one, thanks everyone for the help!

 

[learningphysics]

Homework Statement

Express as a single logarithm 2lnx + ln (x-1) - ln (x-2)

Homework Equations

None used (Basic Algebra)

The Attempt at a Solution

So, let's substitute ln for a, to make it simpler
2ax +a (x-1) - a (x-2)
2ax +ax - a - ax +2a
2ax - a +2a
2ax+a
2lnx+ln
ln (2x+1)

I think that's correct, I'd appreciate some re-check just in case though.

That's the last one, thanks everyone for the help!

You can't substitute for ln... ln is not a variable. it's a function... You could substitute for something like ln(a) which is a value, or ln(x)... but not just ln...

 

[rocomath]

2lnx = ln(x^2)

ln(x-1) - ln(x-2) = ln[(x-1)/(x-2)]

ln(x^2) + ln[(x-1)/(x-2)] = ln[(x^3-x^2)/(x-2)]

 

[matadorqk]

You can't substitute for ln... ln is not a variable. it's a function... You could substitute for something like ln(a) which is a value, or ln(x)... but not just ln...

Uh oh.. Ok let's restart again then.

So ln= log_{e}x

Ok, ill go from there and get back to you

**Or take roco's idea, a head start :)

 

[rocomath]

**Or take roco's idea, a head start :)

nooo!

 

[matadorqk]

nooo!

... I won't just copy it and not learn no worries.
A couple questions:

Is 2lnx=lnx^2 a formula or 'standard'?
Ohh wait, do you get that from the log formulas? That explains a lot!
So you are using \log_{a}x^y=y\log_{a}x

And then you use \log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y..

So that's how you do your first two steps. Then you insert lnx^2 into the ln (x-1) in a reverse distribution. I think I got that correctly, but is that as much as I can simplify?

Would \frac{\ln(x^3-x^2)}{x-2} be the final answer? I stared at it for a while haha, can't find a way to simplify more.

 

[learningphysics]

... I won't just copy it and not learn no worries.
A couple questions:

Is 2lnx=lnx^2 a formula or 'standard'?
Ohh wait, do you get that from the log formulas? That explains a lot!
So you are using \log_{a}x^y=y\log_{a}x

And then you use \log_{a}(\frac{x}{y})=\log_{a}x-\log_{a}y..

So that's how you do your first two steps. Then you insert lnx^2 into the ln (x-1) in a reverse distribution. I think I got that correctly, but is that as much as I can simplify?

Would \frac{\ln(x^3-x^2)}{x-2} be the final answer? I stared at it for a while haha, can't find a way to simplify more.

Careful. the entire fraction should be inside the ln...

 

[matadorqk]

Careful. the entire fraction should be inside the ln...

Right!

ln(\frac{x^3-x^2}{x-2})

 

[learningphysics]

Right!

ln(\frac{x^3-x^2}{x-2})

That's the answer. I don't think you can simplify any more.

 

[matadorqk]

Well, not to take the risk, there's one more problem I did without consulting, so I rather check with you to see if I did it correct. This one I feel a bit more confident about, let's see how that goes:
1. Homework Statement

Solve for x: \log_{3}x + \log_{3}(x-2)=1

2. Homework Equations

#1\log_{a}x+\log_{a}y=\log_{a}xy
#2y=\log_{b}(x)
#3x=b^y
3. The Attempt at a Solution

Ok so let's solve using the formula #1:

So:
\log_{3}(x)(x-2)=1
\log_{3}(x^2-2x)=1
To make this simpler, let's make x^2-2x=w
log_{3}(w)=1
So use formula #2/#3 to get:
w=3^1

x^2-2x=3

Now I kind of got confused on what to do. I seriously don't know why I can't algebraically solve it, by trial and error I got x=3.. but perhaps its the excessive coke or sometihng that's blocking me from algebraically solving this. Help?

 

[rocomath]

have you taken Chemistry II? it's heavily used for Acids/Bases and Reaction Rates, easier than this tho.

 

[matadorqk]

have you taken Chemistry II? it's heavily used for Acids/Bases and Reaction Rates, easier than this tho.

Nope.. I only took General Chemistry a year ago (10th)..

 

[learningphysics]

Have you learned how to solve quadratic equations?

 

[rocomath]

you're already at the 2nd to the last step

x^2 - 2x - 3 = 0

now solve for x

 

[matadorqk]

Have you learned how to solve quadratic equations?

OHH OK, I think I can solve that!

X=3 or x=-1

as (x-3)(x+1)=0

 

[learningphysics]

OHH OK, I think I can solve that!

X=3 or x=-1

as (x-3)(x+1)=0

Yes, but x=-1 is an extraneous root... so your answer is just x = 3.

 

[Feldoh]

To clarify it's extraneous because
\log_{3}(-1) + \log_{3}((-1)-2)=1

However we encounter a problem here since

log_{a}b = c, "b" must be >0

Clearly in this problem it's not and doesn't work.

 

[HallsofIvy]

To clarify it's extraneous because
\log_{3}(-1) + \log_{3}((-1)-2)=1

Well, no, it's not. log₃(-1) and log₃(-3) are undefined and have no value- their sum is NOT equal to 1, it doesn't equal anything.

However we encounter a problem here since

log_{a}b = c, "b" must be >0

Clearly in this problem it's not and doesn't work.

Yes, that's the reason -1 is "extraneous". It satisfies the equation x²- 2x= 3 but not log₃(x)+ log₃(x-1)= 1.

 

[Feldoh]

Well, no, it's not. log₃(-1) and log₃(-3) are undefined and have no value- their sum is NOT equal to 1, it doesn't equal anything.


Yes, that's the reason -1 is "extraneous". It satisfies the equation x²- 2x= 3 but not log₃(x)+ log₃(x-1)= 1.

\log_{3}(-1) + \log_{3}((-1)-2)=1 was just to show an easy way to check -- plug the numbers into the equation. Doesn't match the domain of a log so it doesn't exist...