Topical Revision Paper 5 Unit 5 Exponents & Logarithms

Topical Revision Paper 5 Term 2 Unit 5: Exponents & Logarithms Question 1: (i) Show that   2 333log 1 log 3 log 3( 1)xx    . [4] (ii) Hence, solve the equation    3 33log 1 log 1 log 3xx    . [4] Solution: (i)           33 3 3 3 3 2 3 log 1 log 3 log 3 1 log 3 1 log 3 2log 3 1 log 3 1 (shown) x x x x x      (ii)              3 33 3 33 2 33 2 2 2 log 1 log 1 log 3 log 1 log 1 log 3 log 1 log 3( 1) 1 3( 1) 3( 1) 1 0 3 7 2 0 3 1 2 0 1 (n.a.) or 23 xx xx xx xx xx xx xx xx                          Question 2: Solve the simultaneous equations yx xy 84 2 log2log 7log   [4] Solution:

(1) log3 4log 8log log24log log 22 2 2 2 2 yx yx   (2) 7loglog 22  yx Sub (1) into (2), 8 3log 7log3 7 2 2    y y y 16 4log2   x x Question 3: Solve for x , given that 2 1 2 43 3 3 9 x x x   . [4] Solution: 2 1 2 43 3 3 3 3 9 x x x      Let 3 xu  21 1 4 3 9 9u u u   2 2 3 9 4 3 8 4 0 u u u uu       (3 2)( 2) 0 2 or 23 uu u     32 lg 2 or 0.631lg 3 x x   23 3 2lg 3 or 0.369lg 3 x x   Question 4: Solve the following equations a) 8 2 125log [2 log ( 4)] log 5x   (b) 2ln lg1.6ye 

Solution: (i) 8 2 125 8 2 1log [2 log ( 4)] log 5 log [2 log ( 4)] 3xx       3 28 2 log ( 4) x   2log ( 4) 0x 41x 5x (ii) 2ln lg1.6 2ln ln(lg1.6)yey    ln(lg1.6)ln 2y ln(lg1.6) 2ye 0.452 (3 s.f.)y Question 5: The mass, m grams, of a radioactive substance after t years is given by the formula ktm ae  , where a and k are constants. At the beginning, the mass of radioactive substance is 83 g. After 35 years, there was only 26 g of the radioactive substance left. (i) Find the value of a and of k . [4] (ii) Sketch the graph of m against t. [2] Solution: (i) 083 83 ktm ae ae a   

35 35 26 83 26 83 2635 ln 83 26ln 83 0.033235 k k e e k k       (ii)