logarithmic question

[math321]

can anyone please help with this question

log (3x) = log 4 + log (x - 1)

 

[pka]

That is equivalent to \(\displaystyle \log \left( {\frac{{4x - 4}}{{3x}}} \right) = 0\).

 

[math321]

and then wat
can u jus break it down a little simpler for me please

thank you

 

[lookagain]

That is equivalent to \(\displaystyle \log \left( {\frac{{4x - 4}}{{3x}}} \right) = 0\).

math321,

this is equivalent to \(\displaystyle \log_{10} \left({\frac{{4x - 4}}{3x}}} \right) = 0\)


Change this equation to its equivalent exponential form:

\(\displaystyle 10^0 \ = \ \frac{4x - 4}{3x}\)

\(\displaystyle 1 \ = \ \frac{4x - 4}{3x}\)


Solve for \(\displaystyle x,\) but you must check those candidate solutions in the original equation.