Evaluating log equations

1. Sep 15, 2010

Elus

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

Problem 1:

Solve log(x^3) = (log(x))^2 for x.

Note, there are two solutions, A and B, where A < B.

A = ?
B = ?

Problem 2:

Rewrite the expression 2log(x)-2log(x^2+1)+4log(x-1) as a single logarithm log A.

A = ?

2. The attempt at a solution

Problem 1 Attempt:

I might be able solve this graphically by plotting y1 = log(3^x) and y2 = (log(x))^2 and finding the points of intersection.

However, I'd like to do this by hand, if possible! Would I start by raising both sides to e?

e^log(x^3) = e^(logx)^2

then

x^3 = e^2

then

x = (e^2)^(1/3)
x = e^(2/3)
x = 1.947734

Wolfram-alpha says otherwise :( and there are supposed to be 2 solutions. I only got one D:

Problem 2 attempt:

Okay, I tried using properties of logarithms here.

2log(x)-2log(x^2+1)+4log(x-1)

2log(x) - 8*log(x^3 - x^2 + x - 1) <------- Since the 2 logs are added, you can

multiply the insides of the logs together, right?

16*log(x/(x^3-x^2+x-1)) <------------ Since the 2 logs are subtracted, you can

divide them, right?

Last edited: Sep 15, 2010
2. Sep 15, 2010

Mentallic

For 1) $$e^{log(x)^2}\neq e^2$$
and there's a simpler way to solve this problem anyway, if you let y=log(x) then substituting this into the equation $$log(x^3)=(log(x))^2$$ will give?

For 2)
Well this is like saying $$a.log(b)+c.log(d)=ac.log(bc)$$ this isn't right. If you have a.log(b) then this is equivalent to log(b^a).

Again, not right because you have made the same mistake as before.

3. Sep 15, 2010

Elus

For problem 1:

log(x^3) = (log(x))^2
log(x^3) = y^2 <---Substitute y=log(x) on right side
3*log(x) = y^2 <----- You said log(b^a) = a.log(b), so that's what I applied.
3*y = y^2 <----- substitute y=log(x) on left side
3 = y <----divide both sides by y
3 = log(x) <-----substitute back log(x) for y
x = e^3 <---------raise both sides by e
x = 20.085536

Great, that's one answer, but it said that there is are two answers (A and B). Where do I get the other answer from?

For problem 2:

Okay, so I need to bring the 2, 2, and 4 into the log function?

2log(x)-2log(x^2+1)+4log(x-1)
log(x^2)-log((x^2+1)^2)+(log(x-1)^4)

Before I spend a time solving the ugly problem above, can someone let me know if this going in the right direction?

4. Sep 15, 2010

Mentallic

Yes I'm glad you were able to apply all the formulae correctly
Remember back when you were learning quadratics that when you have something like $$x^2-x=0$$ rather than dividing through by x you factored out x to get $$x(x-1)=0$$ which then gave you the two solutions x=0,1. When you divide through by x you lose the solution x=0 because it is assuming x does not equal 0 (since you can't divide by 0). Do the same for this problem.

Yes that's exactly right. And you don't need to expand, leaving it in factored form is enough when applying the log(a)+log(b)=log(ab) rule.

5. Sep 20, 2010

zgozvrm

Be careful. "log" generally assumes [itex]log_{10}[/tex] whereas "ln" assumes [itex]log_e[/tex]

Last edited: Sep 21, 2010
6. Sep 21, 2010

Mentallic

In maths, the syllabus usually gives the student a chance to work with all types of log bases when starting off, but once they have a firm grip of this they imply all expressions of log are of the base e, unless specifically stated.

7. Sep 21, 2010

zgozvrm

That's why I said "generally!"

Still, I'm betting that the question, as posed, is referring to [itex]log_{10}[/tex] rather than [itex]log_e[/tex] for 3 reasons:
1) the expressions used "log" rather than "ln"
2) scientific calculators have "log" buttons that work with [itex]log_{10}[/tex] and "ln" buttons the work with [itex]log_e[/tex]
3) in this case, using [itex]log_{10}[/tex] results in nice, tidy integer answers

Of course, you can use any base you want and come up with answers that work. But, I'm betting that the point of the problem had more to do with rearranging the equation to solve for X than than anything else.

8. Sep 21, 2010

Mentallic

Yes I noticed you said generally but that's what I'm arguing. In the world outside theoretical maths, it is generally log10 but in the maths classroom you will find that log is considered loge unless otherwise stated.

I myself got so used to this after doing tons of log questions in class that when logs were introduced into a formula in physics I of course made the mistake of using loge rather than log10.

9. Sep 21, 2010

zgozvrm

Apparently, we're arguing the same thing. I simply told the OP to "be careful" because "log" can be interpreted different ways. I don't know what kind of class he is in, or what level (or what country, for that matter), and therefore, I wouldn't assume one base or the other without more information.

10. Sep 21, 2010

eumyang

In high school, however, my experience is that "log" by itself is assumed to be the common logarithm, ie. base 10. "Ln" is used for base e. This is what I've also seen in some Precalculus textbooks. This is high school in the US.

11. Sep 21, 2010

zgozvrm

... which is why I pointed this out to the OP

(Thank you, eumyang!)

12. Sep 21, 2010

Mentallic

I guess this shows how high schools work in different parts of the world. Thanks eumyang.

13. Sep 22, 2010

Tikoonmunci

(That's why I said "generally!")

You actually said general, but that's beside the point.

14. Sep 22, 2010

Mentallic

Well yeah it is, so why did you bring it up :tongue: