What Is the Next Step in Solving This Logarithmic Equation?

[homeworkhelpls]

I started off by using law of logs to divide the logb (6x/18) but i dont know what to do after, please help.

 

[fresh_42]

View attachment 322211
I started off by using law of logs to divide the logb (6x/18) but i dont know what to do after, please help.

If you have transformed ##\log(6x)-\log(18)## to ##\log(6x/18)## then why did you stop? Put in ##x-1## as well.

Btw.: Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

 

[homeworkhelpls]

If you have transformed ##\log(6x)-\log(18)## to ##\log(6x/18)## then why did you stop? Put in ##x-1## as well.

Btw.: Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

i mean i did transform the equation but after idk how to go on

 

[fresh_42]

i mean i did transform the equation but after idk how to go on

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.

 

[symbolipoint]

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) .

 

[mcastillo356]

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}##