Expert: Paul Klarreich - 7/13/2007

Question
QUESTION: So I'm reviewing all of my precal work from 2 years ago and have become stumped on a very easy looking problem.  

How do I solve something like:

Log(4x)=2.  

Would I assume it has a base of 10, so it's 10^2=4x which = 100= 4x which = x=25?

Thanks a lot.  I need to know as I have a math placement exam very soon!

ANSWER: Questioner:   Joe
Category:  Advanced Math
Question:  So I'm reviewing all of my precal work from 2 years ago and have become stumped on a very easy looking problem.  

How do I solve something like:

Log(4x)=2.  

Would I assume it has a base of 10, so it's 10^2=4x which = 100= 4x which = x=25?

Thanks a lot.  I need to know as I have a math placement exam very soon!
...............................
Hi, Joe,

Yes, that looks right.  Usually (at least as of 50 years ago) we wrote:

log (x) for base-10.  [Called Briggsian logarithms in those days.]

ln (x)  for natural logarithms (base is e)  [A.K.A. Naperian logarithms.]

log(sub-b)(x) for any other logarithm.


---------- FOLLOW-UP ----------

QUESTION: Alright, thanks a lot for clearing that up.  Because you're so good, I'm going to ask you a few more questions I came upon as I was reviewing my old work.

Still pertaining to logs, how would I graph:

y=1+log(2x)

How would I solve: 8 x 10^3x = 12.  I know I have to log both sides, but what would I do with log8 X 3xlog10 = log12?  Do I subtract log8 on both sides to get: 3xlog10= log12-log8?  If that's right, am I allowed to then divide by 3log10?  

And one last thing:  How would I find 3 solutions for: z^3 - 1 = 0.  

Thank you for taking the time to help me.

Answer
Questioner:   Joe
Category:  Advanced Math
---------- FOLLOW-UP ----------

QUESTION: Alright, thanks a lot for clearing that up.  Because you're so good, I'm going to ask you a few more questions I came upon as I was reviewing my old work.

Still pertaining to logs, how would I graph:

y=1+log(2x)

How would I solve: 8 x 10^3x = 12.  I know I have to log both sides, but what would I do with log8 X 3xlog10 = log12?  Do I subtract log8 on both sides to get: 3xlog10= log12-log8?  If that's right, am I allowed to then divide by 3log10?  

And one last thing:  How would I find 3 solutions for: z^3 - 1 = 0.  

Thank you for taking the time to help me.
.................
Hi, Joe,

y = 1 + log(2x)

For this, I will assume that  log means ln.  It really makes little difference for the graph.

Write  y = 1 + log 2 + log x,  and apply your usual graph-transformation rules:

That is, take the simplest graph,  y = log x  and apply the rule that says:

If you know the graph of  y = f(x), and want the graph of  y = f(x) + C, elevate the graph vertically by C units.  (Depress it, of course, if C is negative.)

In this case, first draw the graph of  log x  [passes through  (1,0), asymptotic to the negative y-axis, and always rising, but curving to the right, and domain is x > 0]  

Now just raise the whole thing  (1 + log 2) units vertically.
.......................

How would I solve: 8 x 10^3x = 12

I assume you mean   8 (10^3x) = 12

Yes, take the log of both sides, using base 10, but you might as well simplify first.

10^3x = 12/8 = 3/2

Now  3x = log(3/2)  and  x = (1/3) log(3/2), for which you can use your calculator.

If you go by:  log8 * 3xlog10 = log 12, then use the fact that  log 10 = 1.  (Assuming base-10 logarithms, which you can do here.  If YOU are the one taking logs, you get to choose the base.)


Then log 8 * 3x = log 12
3x = log 12 - log 8, which works out to the same thing.

.................................

z^3 - 1 = 0

This has nothing to do with logarithms, but it is part of the standard precalc curriculum.

This is your DeMoivre's theorem stuff.  You write:

z^3 = 1

in the 'polar' form of complex numbers.

z = r exp(i@),  and  

z^3 = r^3 exp(i3@)  << application of DeMoivre's theorem.)

Now you write three different forms for 1:

1 = 1 exp(i 0)
1 = 1 exp(i 360)    << shouldn't use degrees, but it's easier.
1 = 1 exp(i 720)

And write three equations:

z^3 = r^3 exp(i3@) = 1 exp(i 0)
z^3 = r^3 exp(i3@) = 1 exp(i 360)
z^3 = r^3 exp(i3@) = 1 exp(i 720)

giving three solutions:

r = 1,  @ = 0, which is  z = 1
r = 1,  @ = 120, which is  z = 1(cos 120 + i sin 120)
r = 1,  @ = 240, which is  z = 1(cos 240 + i sin 240)

which you rewrite using your trigonometric diagrams.

ALTERNATIVE SOLUTION, still using standard precalc curriculum.

z^3 - 1 = 0

By inspection (a mathematician's term for getting lucky) you can see that  z = 1 solves the equation.  Therefore, by the Factor Theorem (more standard precalc -- this is getting sickening, you are thinking)  the equation can be factored using  (z - 1):

(z - 1)(z^2 + z + 1) = 0

Now solve z^2 + z + 1 = 0, using the quadratic formula.  You will get imaginary roots:
   - 1 +- sqrt(-3)
z = ----------------
         2

   - 1 +- i sqrt(3)
z = ----------------
         2

which is what you would get using the trigonometry.