SOLVING EXPONENTIAL EQUATIONS
Solving Exponential Equations Using Common Base Method
Solve 5x  25
5x  25
5 x  52
x2
Solution set = {2}
Method 2:
Solve 5x  25
5x = 25
log5[5x] = log5[25]
Take log base-5 of both sides.
xlog5(5) = log10(25)/log10(5)
Use Power Rule and change-of-base formula
x = log10(25)/log10(5)
Note: log5(5) = 1
x=2
Solution set is {2}
1
Solving Exponential Equations Using Common Base Method
Solve 5x 
1
25
Note that 5-2 
1
because 52  25
25
Hence,
1
5x 
25
5x  5-2
x  2
Solution set = {-2}
Method 2:
5x = 1/25
5x = 0.04
log5[5x] = log5[0.04]
[x]log5(5) = log10(0.04)/log10(5)
x = log10(0.04)/log10(5)
Take base-5 log of both sides.
Use Power Rule and change-of-base formula
Note: log5(5) = 1
x = -2
Solution set is {-2}
2
Solving Exponential Equations Using Common Base Method
Solve 10 x  0.01
Note: 0.01 
1
 102
100
1 

2
-2
Recall
10

100;
hence
10



100 

Hence,
10 x  102
x  2
Solution set is {-2}
Method 2:
10x = 0.01
log10[10x] = log10[0.01]
[x]log10(10) = log10(0.01)/log10(10)
x = log10(0.01)/log10(10)
Take base-10 log on both sides.
Use Power Rule and change-of-base formula
Note: log10(10) = 1
x = -2
Solution set is {-2}
3
Solving Exponential Equations Using Common Base Method
Solve 6 x  6
Note:
6  2 61  61/ 2
Hence,
6 x  6  61/ 2
1
x
2
Solution set is {1/2}
4
Solving Exponential Equations Using Common Base Method
Solve 7 x  3 7
Note: 3 7  3 71  71/ 3
Hence,
7 x  3 7  71/ 3
1
x
3
Solution set is {1/3}
5
Solving Exponential Equations Using Common Base Method
Solve 8x  1
Note : 80  1
Hence,
8x  1
8 x  80
x0
Solution set is {0}.
6
Solving Exponential Equations Using Common Base Method
Solve 12 x  1
Note : 120  1
Hence,
12 x  120
x 1
Solution set is {1}
7
Solving Exponential Equations Using Common Base Method
Solve 16 x  8
Solution :
16 x  8
2 
4
x
 23
Rewrite 8 as 23 and 16 as 24
2(4)( x )  23
2 4 x  23
Hence, 4x  3
4x 3

4 4
3
x
4
Solution set is {3/4}
8
Solving Exponential Equations Using Common Base Method
Solve 43 x  16
Solution :
43 x  16
43 x  4 2
Rewrite 16 as 42
Hence, 3x  2
3x 2

3 3
2
x
3
Solution set is {2/3}
9
Solving Exponential Equations Using Common Base Method
Solve 16 x1  64
Solution :
16 x1  64
4 
2 x 1
 43
Rewrite 64 as 43 and 16 as 42
4(2)( x1)  43
4 2 x  2  43
Hence, 2x  2  3
2x  2  2  3  2
2x  1
2x 1

2 2
1
x
2
10
Solving Exponential Equations Using Common Base Method
Solve 92 x1  27 x
Solution :
92 x1  27 x
 32 
2 x 1
  33 
x
Rewrite 27 as  33  and 9 as 32
3(2)(2 x1)  3(3)( x )
34 x2  33 x
Hence, 4x  2  3 x
4 x  2  3x  3x  3x
x20
x2202
x  2
11
Solving Exponential Equations Using Common Base Method
Solve
 
2 x2
  64 
8
x
Solution :
 
2 x2
8
81/2 
2 x2
  64 
  82 
x
x
Rewrite 8  81/ 2 and 64 = 82
8(1/2)(2 x2)  8(2)( x )
8 x1  82 x
Hence,
x  1  2x
x  1  x  2x  x
1 x
12
Solve 10x = 3.91
10 x  3.91
log10 10 x   log10  3.91
Take log base-10 of both sides
xlog10 10   log10  3.91
Using Power Rule to move x to the front
x
log10  3.91
log10 10 
Note: log10 10   1
x  0.592176
Solve 3e5x = 189
3e5 x  189
3e5 x 189

3
3
e5 x  63
ln  e5 x   ln  63
5xln  e   ln  63
5 x  ln  63
Note: ln  e   log e (e)  1
5 x ln  63

5
5
x  0.828627
13
Solve 7 x-3 = 500
7
x  3

 500
log 7 7
x  3
  log 500
7
 x  3 log 7  7   log 7  500 
log  500 
 x  3  10
log10  7 
 x  3  3.1936768
Take log base-7 of both sides
Using Power Rule to move (x  3) to the front
x  3  3  3.1936768  3
x  6.1936768
14
Solve 5x-3 = 9x+2
5
x -3
 9
x  2
x -3
x2
log 5 5    log 5 9  
 x - 3 log5 5   x  2  log 5 9
 x  2  log10 9
 x - 3 
log10 5
 x - 3   x  2 1.365212 
Note: log 5 5  1
Note: log10 (9)/log10 (5) = 1.365212
x - 3  1.365212 x  2.730424
x - 3  3  1.365212 x  2.730424  3
x  1.365212 x  5.730424
x  1.365212 x  1.365212 x  1.365212 x  5.730424
0.365212 x  5.730424
0.365212 x 5.730424

0.365212 0.365212
x  15.690678
15
Solve 52x-3 = 43x-1
5
2 x -3
 4
3 x -1
2 x -3
3 x -1
log 5 5    log 5  4  
 2 x  3 log 5 5   3x -1 log 5  4
 3x -1 log10  4
 2 x  3 
log10 5
 2 x  3   3x -1 0.861353
Note: log 5 5  1
Note: log10 (4)/log10 (5) = 0.861353
2 x  3  2.584059 x  0.861353
2 x  3  3  2.584059 x  0.861353  3
2 x  2.584059 x  2.138647
2 x  2.584059 x  2.584059 x  2.584059 x  2.138647
0.584059 x  2.138647
0.584059 x 2.138647

0.584059 0.584059
x  3.661696
16
Solve e 2x - 3e x + 2 = 0
Note: e 2 x   e x 
2
Hence, e 2 x  3e x  2  0   e x   3  e x   2  0
2
If we let u = e x , then  e x   3  e x   2  0  u 2  3u  2  0
2
Solve u 2  3u  2  0
 u  2  u  1  0
u  2; u  1
Thus,
for u  2 :
ex  2
ln e x  ln 2
x ln e  ln 2
x  ln 2 / ln e
x  0.693147
for u  1:
ex  1
ln e x  ln1
x ln e  ln1
x  ln1 / ln e
x0
17
Solve 2e 4 x - 3e 2 x  20 = 0
Note: e 4 x   e 2 x 
2
Hence, 2e 4 x  3e 2 x  20  0  2  e 2 x   3  e 2 x   20  0
2
If we let u = e 2 x ,
then 2  e 2 x   3  e x   20 is the same as 2u 2  3u  20  0
2
Solve 2u 2  3u  20  0
(2u  5)(u - 4)  0
set 2u  5  0;
set u  4  0
2u  - 5
u  4
u  - 5/ 2
Thus,
for u  5/ 2 :
e 2 x  5/ 2
ln e 2 x  ln( 5/ 2)
Since ln( 5/ 2) is undefined, we cannot use -5/2.
for u  4 :
e2 x  4
ln e 2 x  ln 4
2x ln e  ln 4
2x  ln 4
ln 4
x
 0.693147
2
Note: ln e  1
18