how to solve this equation?
log (3x - 2) = 1
thank you
I assume this is a natural logarithm (of base $\displaystyle e$). If not, just replace $\displaystyle e$ with whatever your base happens to be...
$\displaystyle \log{(3x - 2)} = 1$
$\displaystyle e^{\log{(3x - 2)}} = e^1$
$\displaystyle 3x - 2 = e$
$\displaystyle 3x = e + 2$
$\displaystyle x = \frac{e + 2}{3}$.