Solving log₃(3x-5) + log₁/₃(12) = 1-x

[songoku]

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

 

[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.

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

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

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

 

[songoku]

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.

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

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

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

 

[ehild]

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(3^x-5) + log_(1/3) 12 = 1 - x?

ehild

 

[symbolipoint]

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

 

[symbolipoint]

That can be solved only numerically. Are you sure, it was not log₃(3^x-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.

 

[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.

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

ehild

 

[symbolipoint]

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

ehild

?
No.

 

[songoku]

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

ehild

I am sure. I have checked it again.

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 . log_a b, correct?

 

[ehild]

I am sure. I have checked it again.

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


ehild

 

[ehild]

?
No.

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

ehild

 

[symbolipoint]

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.

 

[songoku]

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 ?

 

[symbolipoint]

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.

 

[Saitama]

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

 

[SammyS]

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.

 

[symbolipoint]

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.

 

[songoku]

Thanks for all the help given here