SOLVING LOG EQUATIONS
Review:
1) Express 5 in terms of log base 3
5 = log 3 (35 )  log 3 (243)
2) Express 15 in terms of log base 2
15 = log 2 (215 )  log 2 (32768)
3) Express 4 in terms of log base 3
4 = log 3 (34 )  log 3 (81)
4) Express -1 in terms of log base 8
-1 = log 8 (81 )  log 8 (0.125)
5) Express -2 in terms of log base 4
-2 = log 4 (42 )  log 4 (0.0625)
6) Express -3 in terms of log base e
Note: e  2.718281828
-3 = log e (e 3 )  log e (0.0497870684)  ln(0.0497870684)
7) Express -4 in terms of log base 
-4 = log ( 4 )  log (0.0102659823)
8) Express 5 in terms of log base e
Note: e  2.718281828
5 = log e (e5 )  log e (148.4131591)  ln(148.4131591)
9) Express 3.5 in terms of log base e
Note: e  2.718281828
3.5 = log e (e3.5 )  log e (33.11545196)  ln(33.11545196)
Solve log 5 x = 3
Rewrite 3 in terms of log base-5  3 = log 5 (53 )  log 5 (125)
log 5 x  3
log 5 x  log 5 (125)
x  125
Solution set is {125}
Check answer:
log 5 x  3
log 5 (125)  3
33
Solve log 5 (x  8) = 3
Rewrite constant 3 in terms of log base-5  3 = log 5 (53 )  log 5 (125)
log 5 ( x  8)  3
log 5 ( x  8)  log 5 (125)
( x  8)  125
x  8  8  125  8
x  133
Solution set is {133}
Check answer:
log 5 ( x  8)  3
log 5 (133  8)  3
log 5 (125)  3
log10 (125)
3
log10 (5)
33
Solve ln x  4 = 1
Solution
Rewrite 1 in terms of ln
ln x  4 = 1
ln x  4 = ln(e)
x4 e

x4

2
 e
2
x  4  e2
x  4  4  e2  4
x  e2  4
Solution set is {e 2  4}
 1 = log e ( e1 )  ln(e)