SAJC Check your Understanding_Exponential & Logarithmic Functions_teacher

Check your Understanding (Exponential & Logarithmic Functions) Section 1: Exponential Functions 1. By letting xye= , find the values of x for which 21xxee −−= . [4] 21xxee −−= 2 1y y − = 2 20yy − − = ( )( )2 1 0yy − + = 2 or 1y = − 2 or 1xe = − (reject 1− since 0xe  ) ln 2x= 2. By means of the substitution 212 xu += , or otherwise, solve the equation: 21 21 13 222 x x + + −= [3] ( )( ) 2 2 13 2 2 2 3 2 3 2 0 2 1 2 0 uu uu uu uu −= −= + − = − + = 2 1 1 1 22 122 2 2 1 1 1 x u or u x x +− = =− == + =− =−

3. By using the substitution 3xy= , or otherwise, solve the equation ( ) 11 22 3 9 53 x x +− −= . [5] ( )32 9 3 533 x x  −= Let 3xy= . 32 9 53 yy  −= 29 53 6 0yy+ − = ( )( )9 1 6 0yy− + = 1 9y= or 6y=− So, 13 9 x = or 36x =− (reject) 2x=− 4. Find the exact answer(s) for the equation 3ex = 2e–x – 1. [4] Given 3 2 1xxee −=− , Let xye= , 231y y=− 23 2 0yy+ − = (3 2)( 1) 0yy− + = 2 or 13y=− . So 2 or 1 [N.A.]3 xxee= =− 2ln .3x=

5. Show that 31 13 27 9 y x− =  can be expressed as 3 2 4xy+= . [2] 31 13 27 9 y x− =  3 1 3 23 3 (3 )xy−−= 3 1 3 23 3 (3 )xy−−= 3 1 3 233xy−−= 3 1 3 2xy− = − 3 2 4xy+= (Shown) 6. Find the exact value of x for which 14(2 ) 33 5(2 )xx− += . [5] 14(2 ) 33 5(2 )xx− += Let 2xu= 2 14 33 5 5 33 14 0 (5 2)( 7) 0 uu uu uu += − − = + − = 2 ( ); 75 ln 727 ln 2 x u NA u x =− = =  =

7. Find the exact value of x such that 2e 6 exx=+ . [4] 2 6xxee =+ Let y = ex, y2 – y – 6 = 0 (y + 2)(y – 3) = 0 y = -2 or y = 3 ex = -2 or ex = 3 (Rejected since ex > 0) x = ln3 8. Use an algebraic method to solve the simultaneous equations 2 23 11 32 ,4 3 1. x y xy − + +  = = [5] ( ) ( ) 23 1 2 2 3 5 1 1 324 22 4 6 5 5 5 4 1 .......(1) x y xy xy yx − + − − +  = = − + = + += 2 2 2 31 0 ......(2) xy xy xy + =  + =  =− Sub (2) into (1): ( )( ) 2 2 1 4 1 16 5 4 1 4 5 1 0 4 1 1 0 1or 1 yy yy yy y y x x −= − + = − − = = =  =− =−

9. Solve the simultaneous equations 2x – 5y = 3 and 2x – 3 = 21 – 5y – 2. [4] Let A = 2x , B = 5y A – B = 3 … (1) 420082525218 =+−= BABA …. (2) Solving (1) & (2), A = 128 , B = 125 Hence 2x = 128 = 27  x = 7 and 5y = 125 = 53  y = 3 Section 2: Logarithmic Functions 1. (a) It is given that 9logwx= . Find 1 x , in terms of w, [2] (b) Given that 4ln 3ln ln 2 ln ln54x y x+ − = − , find the value of x y . [4] (a) 9 1log 9 9 www x x x −=  =  = (b) 4ln 3ln ln 2 ln ln54x y x+ − = − 34 ln ln2 54 xy x   =        34 2 54 xy x= 3 33x y = 3x y=

2. Solve the equation: 2 2 23 log log ( 3) log ( 3)x x x+ = + + − [3] ( ) ( ) 2 2 2 22 2 3 log log ( 3) log ( 3) log 8 log ( 3)( 3) 8 ( 3)( 3