- #1
homeworkhelpls
- 41
- 1
I started off by using law of logs to divide the logb (6x/18) but i dont know what to do after, please help.
What Is the Next Step in Solving This Logarithmic Equation?
- Thread starter homeworkhelpls
- Start date
-
- Tags
- Log
Click For Summary
SUMMARY
The discussion focuses on solving the logarithmic equation involving the transformation of the expression logb(6x/18) using logarithmic properties. Participants emphasize the importance of merging terms, specifically log(6x) - log(18) into log(6x/18) and incorporating the term log(x-1). The solution path leads to the equation 3(x+4) = 2x(x-1) through proper application of logarithmic laws. Key logarithmic properties are reiterated to aid in the simplification process.
PREREQUISITES- Understanding of logarithmic properties, including the laws of logarithms.
- Familiarity with algebraic manipulation of equations.
- Basic knowledge of logarithmic functions and their applications.
- Ability to work with equations involving multiple variables.
- Study the properties of logarithms in detail, including the change of base formula.
- Practice solving logarithmic equations with varying bases and complexities.
- Explore algebraic techniques for simplifying expressions involving logarithms.
- Learn how to graph logarithmic functions and interpret their behavior.
Students, educators, and anyone interested in mastering logarithmic equations and their applications in algebra.
I started off by using law of logs to divide the logb (6x/18) but i dont know what to do after, please help.
- #2
- 20,815
- 28,437
If you have transformed ##\log(6x)-\log(18)## to ##\log(6x/18)## then why did you stop? Put in ##x-1## as well.homeworkhelpls said:
Btw.: Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/
- #3
homeworkhelpls
- 41
- 1
i mean i did transform the equation but after idk how to go onfresh_42 said:
- #4
- 20,815
- 28,437
Merge ##\log\left(\dfrac{6x}{18}\right)+\log(x-1)##. Then you get an equation ##\log \ldots = \log \ldots## which you can take ##b## to the power of it.homeworkhelpls said:
- #5
symbolipoint
Homework Helper
Education Advisor
Gold Member
- 7,629
- 2,060
All terms are to the same base b. Properly using the logarithm properties and some simplifications should bring you to a step showing 3(x+4)=2x(x-1) .
- #6
mcastillo356
Gold Member
- 655
- 359
Laws of Logarithms
If ##x>0##, ##y>0##, ##a>0##, ##b>0##, ##a\neq 1##, and ##b\neq 1##, then
(i) ##\log_a 1=0##
(iii)##\log_a {(xy)}=\log_a x+\log_a y##
(iii)##\log_a {\left(\dfrac{1}{x}\right)}=-\log_a x##
(iv)##\log_a {\left(\dfrac{x}{y}\right)}=\log_a x-\log_a y##
(v)##\log_a {(x^y)}=y\log_a x##
(vi)##\log_a x=\displaystyle\frac{\log_b x}{\log_b a}##
If ##x>0##, ##y>0##, ##a>0##, ##b>0##, ##a\neq 1##, and ##b\neq 1##, then
(i) ##\log_a 1=0##
(iii)##\log_a {(xy)}=\log_a x+\log_a y##
(iii)##\log_a {\left(\dfrac{1}{x}\right)}=-\log_a x##
(iv)##\log_a {\left(\dfrac{x}{y}\right)}=\log_a x-\log_a y##
(v)##\log_a {(x^y)}=y\log_a x##
(vi)##\log_a x=\displaystyle\frac{\log_b x}{\log_b a}##