1. ## logartimic equation

how to solve this equation?
log (3x - 2) = 1
thank you

2. Originally Posted by alessandromangione
how to solve this equation?
log (3x - 2) = 1
thank you
I assume this is a natural logarithm (of base $e$). If not, just replace $e$ with whatever your base happens to be...

$\log{(3x - 2)} = 1$

$e^{\log{(3x - 2)}} = e^1$

$3x - 2 = e$

$3x = e + 2$

$x = \frac{e + 2}{3}$.

3. i found another method...but i dont' get this method...can u explain to me? expcially how it got '10'...
3x - 2 = 10 1 . Rewrite this log equation in exponential form. 3x - 2 = 10
3x = 12
x = 4
Solve this linear equation for x.

4. Originally Posted by alessandromangione
i found another method...but i dont' get this method...can u explain to me? expcially how it got '10'...
3x - 2 = 10 1 . Rewrite this log equation in exponential form. 3x - 2 = 10
3x = 12
x = 4
Solve this linear equation for x.
In that case, the base of your logarithm is $10$, not $e$.

Follow the same instructions as I gave you, but replace $e$ with $10$.