- #1
songoku
- 2,291
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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:
[tex]\frac{3x-5}{12}=3^{1-x}[/tex]
Can this be solved??
Thanks
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.
[itex]\[
\log _b x = \frac{{\log _k x}}{{\log _k b}}
\]
[/itex]
[itex]\[
\log _{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}}} 12 = \frac{{\log _3 12}}{{\log _3 \left( {{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}}} \right)}}
\]
[/itex]
Please excuse the slightly bad formatting result in lefthand member of last equation. "Logarithm to the base one-third of twelve..."
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:
[tex]\frac{3x-5}{12}=3^{1-x}[/tex]
Can this be solved??
Thanks
ehild said:That can be solved only numerically. Are you sure, it was not log3(3x-5) + log(1/3) 12 = 1 - x?
ehild
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.
ehild said:Songoku did the transformation correctly. What is log3(1/3) ?
ehild
ehild said:That can be solved only numerically. Are you sure, it was not log3(3x-5) + log(1/3) 12 = 1 - x?
ehild
symbolipoint said:Your second line is wrong. Look again at the change of base formula.
It can be a typo in the book then.songoku said:I am sure. I have checked it again.
symbolipoint said:?
No.
ehild said:Well, how do you transform the equation then? Show your result.
ehild
symbolipoint said:This is shown in post #2 for the treatment of the term that used the rational base of one-third.
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 ?
Using that treatment you also get songoku's result ofsymbolipoint said:This is shown in post #2 for the treatment of the term that used the rational base of one-third.
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:
[tex]\frac{3x-5}{12}=3^{1-x}[/tex]
Can this be solved??
Thanks
A logarithm equation is an equation that involves a logarithm function, which is the inverse of an exponential function. It is typically written in the form of logb(x) = y, where b is the base, x is the argument, and y is the result.
To solve a logarithm equation, you first need to isolate the logarithm on one side of the equation. This can be done by using logarithm properties, such as the power rule and the product rule. Once the logarithm is isolated, you can use the definition of logarithms to rewrite the equation in exponential form and solve for the unknown variable.
The strategy for solving logarithm equations is to first isolate the logarithm, then rewrite the equation in exponential form, and finally solve for the unknown variable. It is important to check the solutions to make sure they are valid and do not result in taking the logarithm of a negative number.
Yes, a logarithm equation can have multiple solutions. This is because the logarithm function is not one-to-one, meaning that different inputs can result in the same output. It is important to check the solutions and make sure they are valid.
It is important to check the solutions of a logarithm equation because the logarithm function is only defined for positive arguments. Therefore, any solutions that result in taking the logarithm of a negative number are not valid. Additionally, some logarithm equations may have extraneous solutions, which can be ruled out by checking the solutions in the original equation.