ACJC JC1 H1 Maths Rev A-1 Complete Solution Exp and Log Functions

2023 JC1 H1 REVISION SET A-1 COMPLETE SOLUTIONS Qn EXPONENTIAL & LOGARITHMIC FUNCTIONS 1(a) VJC 2011 MYE Q2 ( ) 9lg 8 lg 1 lg 24 xx    − + = +       ( ) 9lg 8 lg lg10 lg 42 8 10lg lg 942 4 32 20 9 36 288 20 18 xx x x x x xx x    − − = −          −   =       − = −= = (b) ( ) ( ) 153log12log 55 =−−+ xx ( ) ( ) 153 12log5 =− + x x ( ) ( ) 553 12 =− + x x ( ) ( )53512 −=+ xx 251512 −=+ xx 2613 =x 2=x 2 SRJC 2016 PROMO Q3 (a) ln (pe25) = ln p + ln(e25) = k + 25 ln(e) = k + 25 (b) 3e x – 4e –x = 11 3e2x – 4 = 11ex Let y = ex 3y2 – 4 = 11y 3y2 – 11y – 4 = 0 (3y + 1)(y – 4) = 0 y = 1 3− or y = 4 ex = 1 3− (rejected ex > 0 for all x) or ex = 4 x = ln 4 3 CJC 2012 PROMO Q3 (a) xxx eee −=− 352 3 352 24 =− xx ee 0352 24 =−− xx ee

Qn EXPONENTIAL & LOGARITHMIC FUNCTIONS Let xey 2= . 0352 2 =−− yy 0)3)(12( =−+ yy 2 1−=y or 3=y 2 12 −= xe (rejected, since 02 xe ) or 32 =xe So, 2 3ln=x (b) xaax aaa log2log)(log 2 += xalog41+= 1 4(3) 13= + = 4 RI 2012 PROMO Q 2 x = ln ( y + 1)  y + 1 = e x  y = (1 + y + 1) (1 − y − 1 )  y 2 + 3y = 0  y ( y + 3) = 0  y = 0 or y = −3 When y = −3, x = ln (−2) (N.A.) When y = 0, x = ln (1) = 0 5 36 2 6 42 4264 2 xx yy=  = i.e. 6 6 2 3 3x y y x− =  = − lg( ) 1 lg( 1)y x x− = − − lg(2 3) 1 lg( 1)xx− = − − lg(2 3) lg( 1) 1xx− + − =  lg (2 3)( 1) lg10xx− − = (2 3)( 1) 10xx− − = 22 5 7 0xx− − = (2 7)( 1) 0xx− + = 1x =− or 3.5 Reject 1x =− since lg( 1)x− is undefined. When 3.5x= , from 3 3 10.5 3 7.5yx= − = − = Therefore 3.5x= and 7.5y= 6 TJC 2016 PROMO Q3 For 2 44log log 0q q p− − = 2 2 4 4 4 4log log 0 log log qq q p p q p q− − =  =  = or 4 4 42log log logq q p q p− =  = then 4pq= For 4log 2 log 4ppq q p q=  =  = 44pqp = = (shown)

Qn EXPONENTIAL & LOGARITHMIC FUNCTIONS [G1] Shape and intersections From graph, 2p= or 4 (since p > 0) 16q= or 256 7 JJC 2013 Promo Q4 (i) 45eAtPB=− + Given t = 0, P = 5, ( ) ( ) 0 5 45e 5 45 1 50 (Shown) A B B =− + =+ = Hence, 45e 50AtP=− + Given t = 10, P = 35, ( )10 10 10 10 35 45e 50 15 45e 15e 45 1e 3 110 ln 3 10 ln 3 A A A A A A =− + − =− −= − = =  =− 11ln 3 where 10 10Ak=− =− (ii) ( )1 ln31045e 50 t P − =− + When t = 20, ( )( )1 ln3 201045e 50 45P − =− + = Population = 45 000 (iii) ( )1 ln31045e 50 t P − =− + After a very long time, ( )1 ln310e0 t− → , 50P→ . Population = 50 000 (iv) Sketch ( )1 ln31045e 50 t P − =− + P t O 5 50 P = 50

Qn EXPONENTIAL & LOGARITHMIC FUNCTIONS 8 RVHS 2016 PROMO Q11 (i) ( ) ( )( ) When 0, 20 20 ln 0 1 20 When 2, 65 65 20 ln 2 2 1 45 ln(5) 45 ln 5 t bk b t k k k   == = + + = == = + + = = (ii) ( ) ( ) ( ) 2 ln53 2 ln53 When 50, 4550 20 ln 2 1ln 5 4530 ln 2 1ln 5 2 ln 5 ln 2 13 21 1 0.962 minutes (3sf)2