Solve 3 + 2 ln ( x /7+3 ) = -4 .

Solution

This initial step in this problem is to get the logarithm by itself on one side of the equation  along with a coefficient of 1.

2 ln ( x/7 + 3 ) = -7

ln ( x/7 + 3 ) = - 7/2

Now, we have to get the x out of the logarithm and the best way to do i.e. to "exponentiate" both sides using e.  In other terms,

e ^{ln ( x/7 +3)}          =e^{- 7/2}

So using the property above with e, since there is a natural logarithm in the equation, we get,

 (x/7)+3=e^{- 7/2}

Now all that we need to do is solve this for x.

x /7+ 3 = e^{- 7/2}

x/7 =  -3 + e ^{- 7/2}

x = 7 ( -3 + e ^-7/2 ) = -20.78861832

At this instance we might be tempted to say that we're done & move on.  Though, we do have to be careful.  Recall from the earlier section that we can't plug a -ve number into a logarithm.  It, by itself, doesn't mean that our answer won't work as its negative. What we have to do is plug it into the logarithm & ensure that x/7 + 3 will not be negative.  We'll leave it to you to check out that this is actually positive upon plugging our solution into the logarithm and hence x = -20.78861832 is actually a solution to the equation.

 Now let's take a look at a more complexes equation. Often there will be more than one logarithm into the equation. While this happens we will have to use on or more of the following properties to combine all of the logarithms into a single logarithm.  Once it has been done we can proceed as we did in the earlier example.

log _bxy = log _bx + log_b y        log_b( x /y)= log_bx - log_b y          log_b (x^r ) = r log_b x