# Solve for "a" if log (2a-3) - 2 = log (a+3)

### 2 Answers | Add Yours

We have to solve for "a" if log (2a-3) - 2 = log (a+3)

log (2a-3) - 2 = log (a+3)

=> log ( 2a - 3) - log( a + 3) = 2

=> log[(2a - 3)/(a + 3)] = 2

=> (2a - 3)/(a + 3) = 100

=> 2a - 3 = 100a + 300

=> 98a = -303

=> a = -303/ 98

**As log (2a -3) is not defined for a = -303/98, we cannot find any solution**

Given the logarithm equation:

log (2a-3) -2 = log (a+3)

We need to solve for "a"

First we will combine similar terms.

==> log (2a-3) - log (a+3) = 2

Now we will use the logarithm properties to solve.

We know that log a - log b = log a/b

==> log (2a-3)/(a+3) = 2

Now we will rewrite into the exponent form.

==> (2a-3)/(a+3) = 10^2

==> (2a-3)/(a+3) = 100

Now we will multiply by (a+3)

==> 2a -3 = 100(a+3)

==> 2a -3 = 100a +300

==> 98a = -303

**==> a = -303/98 = -3.01**

But the values of a is not defined for log (2a-3) and log 9a+3)

**Then, the equation has no solution.**