Page 85 - Elementary Algebra Exercise Book I
P. 85

ELEMENTARY ALGEBRA EXERCISE BOOK I                                               equAtions




                                                                1
                                                        1
               If a> 0 but  a  =1, then x> 0,x  =1,x  = ,x  = , and log a =     1  , log a =     log a   =    1   , log 2 a =     log a   =     1
                                                                                                   a
                                                                                                                                    a
                                                        a      a 2      x     log x    ax     log a+log x   1+log x   a x     2 log a+log x  2+log x
                                                                                                                                       a
                                                                                                                a
                                                                                                                                 a
                                                                                                                                                 a
                                                                                                      a
                                                                                                a
                                                                                 a
                                      log a
     log a
 1
                   1
                                                    1
 log a =  log x , log a =  log a+log x  =  1+log x  , log 2 a =  2 log a+log x  =  2+log x  . The original equation is equivalent to
       a
                                        a
 x
 ax
                           a x
                                                     a
                                      a
                    a
    a
          a
                                           a
 a
                                                                                             2
                 2   +    1   +     3   =0 . Let  t = log x , then    2    1     3          6t +11t+4          2
                                                                                            2
                                                                                                     =0 ⇒ 6t + 11t +4 = 0 ⇒ (3t + 4)(2t + 1) = 0 ⇒
                                                                                                             2
                                                                                    =0 ⇒ 6t +11t+4
               log x   1+log x   2+log x                  a         2  t + + 1 1+t + + 3 2+t =0 ⇒  t(1+t)(2+t) =0 ⇒ 6t + 11t +4 = 0 ⇒ (3t + 4)(2t + 1) = 0 ⇒
                           a
                  a
                                     a
                                                                    t   4 1+t  2+t        t(1+t)(2+t)
   2
 2  +  1  +  3  =0 ⇒  6t +11t+4  =0 ⇒ 6t + 11t +4 = 0 ⇒ (3t + 4)(2t + 1) = 0 ⇒ t = −  or  t = − . When  t = − , then
                                                                  t = − 4
                     2
                                                                                  1
                                                                                                   4
                                                                        3
 t  1+t  2+t  t(1+t)(2+t)                                             3           2                3
 4
 t = −
 3                        4           4               1                   1           1
               log x = − ⇒ x = a     − 3 . When  t = − , then  log x = − ⇒ x = a    − 2 . It is not difficult
                  a
                                                                  a
                                                                          2
                          3
                                                      2
                                     4        1
               to verify that  x = a − 3 ,x = a − 2  are roots of the original equation.
                                                                                                     n
               2.94      The coefficients of the last three terms of the expansion of  (x    lg x  + 1)  are
                                                     2
                                                    y
               positive  integer roots of the equation 3 · 9 −10y  · 81 −11  =1. The middle term of the expansion
                                                            √

               is the root of the equation  3  m  =0.1 −2  +  2m , find the value of  x .
                                               2


















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