SOLVING LOG EQUATIONS
Solve log 5 x = 3
Rewrite log 5 x  3 in exponential form:
log 5 x  3  53  x
Hence x  125
Solve log 5 (x  8) = 3
Rewrite log 5 ( x  8)  3 in exponential form:
log 5 ( x  8)  3  53  ( x  8)
Hence ,
( x  8)  125
x  8  8  125  8
x  133
Solve ln x  4 = 1
ln x  4 = 1
ln  x  4 
1/2
=1
1 / 2  ln( x  4)  1
1 / 2  ln( x  4)  1
1 / 2 
1 / 2 
ln( x  4)  2
log e ( x  4)  2
e2  x  4
e2  4  x
x  e2  4
Note:
x4 
2
 x  4
1
  x  4
1/2
Solve 3log 4 ( x  4) = 7  log 4 9
3log 4 ( x  4) = 7  log 4 9
log 4 ( x  4)3 = 7  log 4 9
log 4 ( x  4)3  log 4 9 = 7  log 4 9  log 4 9
log 4 ( x  4)3  log 4 9 = 7
log 4 9  ( x  4)3  = 7
47 = 9  ( x  4)3
16384  9  ( x  4)3
16384 9  ( x  4)3

9
9
1820.4444444  ( x  4)3
3
1820.4444444  3 ( x  4)3
12.0210285  x  4
12.0210285  4  x  4  4
x  16.0210285
Solve log 3 (x - 3) + log 3 (x) - log 3 (x + 4) = 5
log 3 ( x  3)  log 3 ( x)  log 3 ( x  4)  5
 ( x  3)( x) 
log 3 
 5
(
x

4)


( x  3)( x)
35 
( x  4)
( x  3)( x)
243 
( x  4)
( x  3)( x)
243( x  4) 
( x  4)
( x  4)
243( x  4)  ( x  3)( x)
243 x  972  x 2  3 x
243 x  972  243 x  972  x 2  3 x  243 x  972
0  x 2  246 x  972
Using the quadratic formula to solve 0  x 2  246 x  972,
2
b  b 2  4ac (246)  (246)  4(1)(972)
x

2a
2(1)
x = 249.88971589534 ;
x = -3.88971589533961;
Note that x = -3.88971589533961 is an extraneous solution.
Thus, only x = 249.88971589534 is the acceptable solution.
Solve 3log x  log125
3log x  log125
log x 3  log125
x 3  125
x 3  3 125
x5
3
Solve log x  log( x  4)  log125
log x  log( x  4)  log125
log  x  x  4    log125
x  x  4   125
x 2  4 x  125
x 2  4 x  125  125  125
x 2  4 x  125  0
Using the quadratic formula to solve x 2  4 x  125  0
2
b  b 2  4ac (4)  (4)  4(1)(125)
x

2a
2(1)
x  13.3578166916005 ;
x  -9.35781669160055;
Note that x  -9.35781669160055 is an extraneous solution.
Thus only x  13.3578166916005 is the acceptable solution.
Solve log x  log( x  4)  log12  log(2 x  14)
log x  log( x  4)  log12  log(2 x  14)
log  x( x  4)   log 12(2 x  14) 
x( x  4)  12(2 x  14)
x 2  4 x  24 x  168
x 2  4 x  24 x  168  24 x  168  24 x  168
x 2  28 x  168  0
Using the quadratic formula to solve x 2  28 x  168  0
2
b  b 2  4ac (28)  (28)  4(1)(168)
x

2a
2(1)
x = 19.2915026221292;
x = 8.70849737787082;