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File: Solving Equations Pdf 178517 | Solving Logarithmic Equations
solving logarithmic equations example 1 write each equation in exponential form 3 1 a log32 8 b log81 3 5 4 2 1 the base is 32 and the exponent ...

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                                                        Solving Logarithmic Equations 
                       
                      Example 1 
                      Write each equation in exponential form. 
                       
                                       3                                                                 1
                      a.  log32 8 =                                                     b.  log81 3 =   
                                       5                                                                 4
                                                                          2                                                                1
                           The base is 32, and the exponent is  .                            The base is 81, and the exponent is  . 
                                                                          3                                                                4
                                    3                                                                 1
                           8 =      5                                                        3 =      4  
                                 32                                                               81
                                                                                                                                                       
                       
                       
                      Example 2 
                      Write each equation in logarithmic form. 
                       
                             4                                                                       1
                      a.  6  = 1296                                                            -8
                           The base is 6, and the exponent, or                          b.  2  = 256 
                           logarithm, is 4.                                                  The base is 2, and the exponent, or 
                           log  1296 = 4                                                     logarithm, is -8. 
                               6
                                                                                             log   1  = -8 
                                                                                                 2 256
                                                                                                                                                       
                       
                       
                      Example 3 
                      Evaluate the expression log   1 . 
                                                            3 27
                       
                      Let x = log   1 . 
                                    3 27
                        x = log   1  
                                  3 27
                         x     1
                        3   =                Definition of logarithm 
                              27
                         x         -1         -m     1
                        3   = (27)           a  =       
                                                      m
                                                    a
                         x       3 -1         3
                        3   = (3 )           3  = 27 
                         x      -3             m n      mn
                        3   = 3              (a )  = a  
                                                 u      v
                        x = -3               If a  = a , then u = v. 
                                                                                                                                                       
                            
                           Example 4 
                           CHEMISTRY Refer to the application at the beginning of Lesson 11-4 in your book. How long 
                           would it take for 640,000 grams of Polonium-194, with a half-life of 0.5 second, to decay to 5000 
                           grams? 
                            
                                                      1 t                                                        1
                                                                                                  t
                                        N = N                               N = N (1 + r) for r = -  
                                                   0 2                                0                          2
                                                               1 t
                                                                                                                    
                                   5000 = 640,000                           N = 5000, N  = 640,000
                                                               2                                0
                                       1          1 t
                                             =                              Divide each side by 640,000. 
                                     128          2
                             log      1  = t                                Write the equation in logarithmic form. 
                                  1 128
                                  2
                              log      1  = t                               128 = 27 
                                    1 27
                                    2
                                     1 7                                                     n
                                                                               11
                                             = t                                                
                            log1 2                                            bn         b
                                 2
                                     1 7          1 t
                                     2  =  2                                Definition of logarithm 
                                         7 = t 
                            
                           It will take 7 half-lives or 3.5 seconds. 
                                                                                                                                                                                        
                      
                     Example 5 
                     Solve each equation. 
                      
                                       1    1
                     a.  logp 65614  =                                               b.  log5 -(5x - 3) = log5 -(10x + 2) 
                                            2
                                       1     1                                           log  -(5x - 3) = log  -(10x + 2) 
                          log          4   =                                                 5                   5
                              p 6561         2                                                 -(5x - 3) = -(10x + 2) 
                                       1           1                                                  5x = -5 
                                     p2  =         4  
                                             6561                                                       x = -1 
                                       p =  4 6561 
                                        2       2
                                  (  p)  = (9)  
                                       p = 81 
                      
                     c.   log8 (x + 1) + log8 (x + 3) = log8 24 
                           log  (x + 1) + log  (x + 3) = log  24 
                               8                 8                 8
                                  log  [(x + 1)(x + 3)] = log  24 
                                      8      x2 + 4x + 3 = 24 8
                                            x2 + 4x - 21 = 0 
                                           (x + 7)(x - 3) = 0 
                            x + 7 = 0           or             x - 3 = 0 
                                 x = -7                             x = 3 
                      
                          By substituting x = -7 and x = 3 into the equation, we find that x = -7 is undefined for the equation 
                          log  (x + 1) + log  (x + 3) = log  24. When x = -7 we get an extraneous solution. 
                              8                 8                 8
                          So, x = 3 is the correct solution. 
                                                                                                                                                  
                      
                      
                     Example 6 
                     Graph y = log4 (x + 2). 
                      
                                                                                  y
                     The equation y = log  (x + 2) can be written as 4  = x + 2. Choose values for y and then find the 
                                               4
                     corresponding values of x. 
                      
                         y         x + 2             x               (x, y) 
                        -3         0.016         -1.984         (-1.984, -3) 
                        -2         0.063         -1.937         (-1.937, -2) 
                        -1          0.25          -1.75          (-1.75, -1) 
                         0            1             -1              (-1, 0) 
                         1            4              2               (2, 1) 
                         2           16             14              (14, 2)                                                       
                         3           64             62              (62, 3) 
                                                    
                                                                                                                                                  
                   
                  Example 7 
                  Graph y ≥ log3 x - 4. 
                   
                  The boundary for the inequality y ≥ log  x - 4 can be written as y = log  x - 4. Rewrite this equation in 
                  exponential form.                         3                                3
                   
                           y       = log  x - 4 
                                         3
                           y + 4   = log  x 
                            y + 4        3
                           3       = x 
                   
                  Use a table of values to graph the boundary. 
                   
                     y        y + 4          x            (x, y) 
                    -7          -3        0.037        (0.037, -7) 
                    -6          -2        0.111        (0.111, -6) 
                    -5          -1        0.333        (0.333, -5) 
                    -4          0            1           (1, -4) 
                    -3          1            3           (3, -3) 
                    -2          2            9           (9, -2)                                              
                    -1          3           27           (27, -1) 
                                            
                   
                  Test a point, for example (0, 0), to determine which region to shade. 
                   
                   y + 4        0 + 4
                  3     ≥ 0     3    ≥ 0  True 
                   
                  Shade the region that contains the point at (0, 0). However, since values of x ≤ 0 yield extraneous 
                  solutions, only shade above the curve in quadrants I and IV. 
                                                                                                                           
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