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)$\log_a x=\displaystyle\frac{\log_b x}{\log_b a}$