How do I solve log 3x = log 2 + log (x+5)

In words it is log base ten of 3x equals log base ten of two plus log base ten of x + 5.

log 3x = log 2 + log (x+5)
log (x+5) = log 3x - log 2 = log(3x/2)
(x+5) = (3x/2)
2x +10= 3x
x = 10

If you have logs with the same base, you can combine them.
Here is how: Log(basex)A + Log(basex)B = Log(basex)AB
and like this also: Log(basex)A - Log(basex)B = Log(basex)(A/B)

log(3x) = log2 + log(x + 5)
log(3x) = log(2(x + 5))
log(3x) = log(2x + 10)
and since they have the same base, you can say this:
3x = 2x + 10
x = 10

If we have 2 logarithmic functions with the same base, then they obey several "log laws". One of these is:
log A + log B = log AB

Hence, for your equation:

log 2 + log (x+5) = log (2x+10)

According to the equation, this also equals log 3x. Therefore:
log 3x = log (2x+10)
3x = 2x + 10
x = 10.

Testing it out:
log (3x10) = log 30 = 3.4012
log 2 + log (10+5) = log 2 + log 15 = 0.6931 + 2.7081 = 3.4012

Hence it is true, x=10.

this magical device called the calculator...

lg3x=lg2+lg(x+5)
lg3x-lg2=lg(x+5)
lg3x/2=lg(x+5)
3x/2=x+5
3x=2x + 10
x=10 :D

log(3x) = log(2) + log (x+5)
log(3x) = log(2(x+5))
3x = 2(x+5)
3x = 2x + 10
x = 10

log 3x - log ( x + 5 ) = log 2

log [ (3x) / ( x + 5 ) ] = log 2

3x / ( x + 5 ) = 2

3x = 2x + 10

x = 10