# Logarithm equation

1. Nov 8, 2012

### songoku

1. The problem statement, all variables and given/known data
log3(3x-5) + log(1/3) 12 = 1 - x

2. Relevant equations
logarithm
exponential

3. The attempt at a solution
The best I can get is:

$$\frac{3x-5}{12}=3^{1-x}$$

Can this be solved??

Thanks

2. Nov 8, 2012

### symbolipoint

You have two different bases and what you did to get your equation in part three ignored the two different bases. Use the change-of-base formula.

$$\log _b x = \frac{{\log _k x}}{{\log _k b}}$$

$$\log _{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 3}}} 12 = \frac{{\log _3 12}}{{\log _3 \left( {{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 3}}} \right)}}$$

Please excuse the slightly bad formatting result in lefthand member of last equation. "Logarithm to the base one-third of twelve...."

Last edited: Nov 8, 2012
3. Nov 8, 2012

### songoku

This is what I did:

log3(3x-5) + log(1/3) 12 = 1 - x

log3(3x-5) + log(3-1)12 = 1 - x

log3(3x-5) - log312 = 1 - x

log3 [(3x-5)/12] = 1 - x

and the last line is the same as what I post in #1

If I understand it correctly, I think I will come up with the same equation as mine using your idea. Am I correct?

Thanks

4. Nov 9, 2012

### ehild

That can be solved only numerically. Are you sure, it was not log3(3x-5) + log(1/3) 12 = 1 - x?

ehild

5. Nov 9, 2012

### symbolipoint

Your second line is wrong. Look again at the change of base formula.

6. Nov 9, 2012

### symbolipoint

songoku did not say if he is trying to solve the equation or not. If so, you're correct, but he may still want a transformed version of the original logarithm equation.

... In fact, the transformed equation in exponential form can be simplified significantly, still having an x in an exponent and and x term in a polynomial.

Last edited: Nov 9, 2012
7. Nov 9, 2012

### ehild

Songoku did the transformation correctly. What is log3(1/3) ?

ehild

8. Nov 9, 2012

?
No.

9. Nov 9, 2012

### songoku

I am sure. I have checked it again.

Why is it wrong?

1/3 = 3-1

and if I have log(an)b I can change it to 1/n . loga b, correct?

10. Nov 9, 2012

### ehild

It can be a typo in the book then.
You can find approximate solution graphically.

ehild

11. Nov 9, 2012

### ehild

Well, how do you transform the equation then? Show your result.

ehild

12. Nov 9, 2012

### symbolipoint

This is shown in post #2 for the treatment of the term that used the rational base of one-third.

13. Nov 9, 2012

### songoku

OK let me test it

log3(3x-5) + log(1/3) 12 = 1 - x

Use change of base rule, this becomes:

log3(3x-5) + [log3 12] / [log3 (1/3)]= 1 - x

log3(3x-5) + [log3 12] / (-1)= 1 - x

log3(3x-5) - [log3 12] = 1 - x ---> same as what I got ???

14. Nov 9, 2012

### symbolipoint

songoku, okay you convinced me. Although you did not quite finish, the work looks good.

15. Nov 9, 2012

### Saitama

ehild is correct, there can be a typo in the book. Plug the equation in wolframalpha.

16. Nov 9, 2012

### SammyS

Staff Emeritus
Using that treatment you also get songoku's result of
$\displaystyle \frac{3x-5}{12}=3^{1-x}$​
which is correct, as ehild has pointed out more than once.

17. Nov 9, 2012

### symbolipoint

I made a small but very significant arithmetic mistake when I made the equation transformation myself the first time. The equation result which you show is good.

18. Nov 9, 2012

### songoku

Thanks for all the help given here