Solving Logarithm Equation: 3x-5/12=3^1-x

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Homework Help Overview

The discussion revolves around solving a logarithmic equation involving different bases: log3(3x-5) + log(1/3) 12 = 1 - x. Participants explore the implications of using the change-of-base formula and the transformations of logarithmic expressions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of the original logarithmic equation and the use of the change-of-base formula. Some question the correctness of the transformations and whether the equation can be solved numerically or if a transformed version is sufficient.

Discussion Status

The discussion includes various interpretations of the logarithmic equation and its transformations. Some participants express confidence in their approaches, while others suggest checking for potential errors and typos in the original problem statement. There is a mix of agreement and differing opinions on the methods used.

Contextual Notes

Participants note the possibility of a typo in the original problem statement and the implications this has for the solution process. There is also mention of the need for numerical solutions and graphical methods to approximate solutions.

songoku
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Homework Statement


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


Homework Equations


logarithm
exponential


The Attempt at a Solution


The best I can get is:

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

Can this be solved??

Thanks
 
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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.

\[<br /> \log _b x = \frac{{\log _k x}}{{\log _k b}}<br /> \]<br />

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

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

\[<br /> \log _b x = \frac{{\log _k x}}{{\log _k b}}<br /> \]<br />

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

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

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 #1If I understand it correctly, I think I will come up with the same equation as mine using your idea. Am I correct?

Thanks
 
songoku said:

Homework Statement


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


Homework Equations


logarithm
exponential


The Attempt at a Solution


The best I can get is:

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

Can this be solved??

Thanks

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

ehild
 
Your second line is wrong. Look again at the change of base formula.
 
ehild said:
That can be solved only numerically. Are you sure, it was not log3(3x-5) + log(1/3) 12 = 1 - x?

ehild

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:
symbolipoint said:
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.

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

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

ehild

?
No.
 
ehild said:
That can be solved only numerically. Are you sure, it was not log3(3x-5) + log(1/3) 12 = 1 - x?

ehild

I am sure. I have checked it again.

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

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
songoku said:
I am sure. I have checked it again.
It can be a typo in the book then.
You can find approximate solution graphically.


ehild
 
  • #11
symbolipoint said:
?
No.

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

ehild
 
  • #12
ehild said:
Well, how do you transform the equation then? Show your result.

ehild

This is shown in post #2 for the treatment of the term that used the rational base of one-third.
 
  • #13
symbolipoint said:
This is shown in post #2 for the treatment of the term that used the rational base of one-third.

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
songoku said:
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 ?

songoku, okay you convinced me. Although you did not quite finish, the work looks good.
 
  • #15
ehild is correct, there can be a typo in the book. Plug the equation in wolframalpha.
 
  • #16
symbolipoint said:
This is shown in post #2 for the treatment of the term that used the rational base of one-third.
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
songoku said:

Homework Statement


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


Homework Equations


logarithm
exponential


The Attempt at a Solution


The best I can get is:

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

Can this be solved??

Thanks

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
Thanks for all the help given here :smile:
 

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