MATHS NOTES Th Institut of Education 06 SUBJECT: Laving Crt Maths LEVEL: Highr Lvl TEACHER: Aidan Roantr Topics Covrd: Algbra (Eponntial and Log Functions) About Aidan: Aidan is th snior Maths tachr at Th Institut of Education, whr h has bn taching Maths and Applid Maths for ovr 6 yars. As th author of 4 books, including th most rcnt Effctiv Maths Books &, h is acquaintd with vry dtail of th nw Maths cours. Aidan has also appard on radio programms to giv advic to studnts prior to ams. H is rgularly calld upon by branchs of th Irish Mathmatics Tachrs Association to giv talks to tachrs on aspcts of th Maths courss.
Unit Algbra 8 Powrs and Logs 8.5 Eponntial Functions and Rlationships A Eponntial Functions and Rlationships. Dfinition of an Eponntial Function * An ponntial function is a function of th form f : a( b), whr a, b, b 0. * For ampl, f ( ) 4(3) and f ( ) 86( 03) ar both ponntial functions. * Thr ar many ral-lif ampls of ponntial functions,.g. th growth of an invstmnt undr compound intrst, th dcay of a radioactiv substanc. Pag
. Eponntial Rlationship from a Data Tabl Th following tabl givs corrsponding valus for two variabls and y. 0 3 4 y 50 65 85 9965 965 * As th changs in th indpndnt variabl,, ar fid, w can chck if thr is an ponntial rlationship btwn and y by calculating th ratios of succssiv trms. If ths ar constant, thn thr is an ponntial rlationship btwn and y, i.. y is an ponntial function of. * Chcking, (i) 65 50, (ii) 85, (iii) 65 9965, (iv) 85 965. 9965 As ths ratios ar constant, y is an ponntial function of. Dfinition of an Eponntial Rlationship Th data in a data tabl rprsnts an ponntial rlationship btwn th variabls if (i) for fid changs in th indpndnt variabl, (ii) th ratios of succssiv trms is constant. 3. Finding th Equation of an Eponntial Rlationship from a Data Tabl * If thr is an ponntial rlationship btwn and y, w can writ y a( b). * W can thn us two data pairs from th tabl to valuat a and b. Eampl Th tabl blow givs corrsponding valus for th variabls and y. 0 3 4 y 50 65 85 9965 965 If thr is an ponntial rlationship btwn and y, find th valus of th constants a and b if y a( b). Solution [] 0 whn y 50. 0 50 a( b) 50 a... as 0 b Pag
[] whn y 65. 65 50( b) 65 50b b Thus y 50( ). (Not that b, th bas of th ponntial function, is always th sam as th constant ratio, as long as th diffrncs in th valus is.) B Eponntial Problms. Eponntial Chang in Ral Lif A quantity y changs ponntially with rspct to anothr quantity if th rat of chang of y is proportional to th currnt valu of. For ampl, * ach hour, a biological sampl incrass by 5% of its siz at th bginning of that hour, * ach yar, a quantity of radioactiv substanc dcras by % of th quantity prsnt at th bginning of that yar.. Eponntial Growth * If a quantity y incrass ponntially as anothr quantity incrass, thn w can writ y a( b), whr b. * Th bas of th powr bing gratr than is th ky to ponntial growth. In practical problms, it is takn that a 0. 3. Eponntial Dcay * If a quantity y dcrass ponntially as anothr quantity incrass, thn w can writ y a( b), whr b. * Th bas of th powr bing lss than is th ky to ponntial dcay. In practical problms, it is takn that a 0. 4. Finding an Ind Logs can b usd to valuat an ind. For ampl, if 67848 (35) thn 330857 35 and log35 330857 3 089, corrct to four dcimal placs. Pag 3
Eampl A biological sampl starts with a siz of P and grows at th rat of 3% ach day aftr that. Lt f ( ) rprsnt th siz of th sampl days latr. (i) Dtrmin what factor P is multiplid by aftr a priod of on day. (ii) Eprss f ( ) in trms of. (iii) Find, corrct to th narst hour, how long it taks for th sampl to doubl in siz. (iv) Find an prssion for in trms of n if f ( ) np. Solution (i) At th nd of th first day, th siz of th sampl is f () P 3% of P f () P 0 03P f () P( 0 03) f () P( 03) Thus th siz of th sampl is multiplid by th factor 03 ovr a priod of on day. (ii) Hnc f () f () ( 03) P(03) (03) P( 03) and f ( ) P( 03) (iii) (W want to find th valu of for which f ( ) P, i.. whn th sampl siz is twic its original siz.) f ( ) P P(03) P (03) R-writing this quation in log form, log 03 3 45 days... using th log function on a calculator 3 days and hours, corrct to th narst hour. (iv) Givn f ( ) np P( 03) np ( 03) n n. log 03 Pag 4
Erciss 8.5. Th tabl blow givs corrsponding valus for th variabls and y. 0 3 4 y 00 05 05 5765 55065 (i) Us th data in th tabl to show that thr is an ponntial rlationship btwn and y. (ii) If y a( b), find th valus of th constants a and b.. Th tabl blow givs corrsponding valus for th variabls and y. 0 4 6 8 y 500 480 4608 44368 446738 (i) Us th data in th tabl to show that thr is an ponntial rlationship btwn and y. (ii) If y a( b), find th valus of th constants a and b. 3. A laboratory obtains a 0 kg sampl of a radioactiv substanc, which dcays ovr tim. Th following masurmnts of wight, W, in grams, wr obtaind on a yarly basis. Lt rprsnt th numbr of yars from whn th substanc was obtaind. / yars 0 3 4 W / grams 0000 9000 800 790 656 (i) Vrify that W is an ponntial function of. (ii) Eprss W in trms of. (iii) Us your prssion for W to calculat th wight of radioactiv substanc prsnt aftr 75 yars. (iv) Us logs to find th numbr of yars it will tak for th amount of radioactiv substanc to rduc to half th original amount. (This is calld th half lif of th substanc.) Giv your answr in yars to thr dcimal placs. (v) Dtrmin th numbr of yars, corrct to thr dcimal placs, that it will tak for th wight of radioactiv substanc to rduc to kg. 4. Each yar 4% of th quantity of a radioactiv substanc prsnt dcays. Th amount of radioactiv substanc prsnt at th start was 500 grams. Lt y f ( ) b th amount of radioactiv substanc prsnt aftr yars. (i) Writ down an prssion for f (), th amount prsnt aftr on yar. (ii) By writing down prssions for f () and f (3), find an prssion for f ( ). (iii) Find th numbr of yars it taks for th amount of radioactiv substanc to rduc to 00 grams. Giv your answr corrct to th narst yar. Pag 5
5. Th numbr of bactria prsnt in a cultur incrass by 3% ach hour. Th numbr of bactria prsnt initially is N. (i) If f ( t ) rprsnts th numbr of bactria prsnt aftr t hours, show that f ( t) N( 03) t. (ii) Find th numbr of hours it taks for th numbr of bactria prsnt to incras to N. Giv your answr corrct to th narst hour. (iii) If th numbr prsnt aftr 6 hours is calculatd to b 4386, find th valu of N corrct to th narst unit. 6. A sum of mony P, invstd in a financial institution, grows by 4 5% ach yar aftr that. Lt A( t ) b th amount to which th sum has grown aftr t yars. (i) Find an prssion for A( t ). (ii) Find t, th valu of t for which th sum of mony has grown by 50%. (iii) Invstigat if A( t) P. (iv) If A( t t) P, find th valu of t. 7. A nw machin costs 00,000. Its valu dprciats by 8% pr yar aftr that. Lt f ( t ) b th valu of th machin t yars aftr bing purchasd. (i) Eprss f ( t ) in th form a( b ) t. (ii) Find th numbr of yars, corrct to two dcimal placs, at which th valu of th machin is half its original valu. (iii) Th company that buys th machin plans on rplacing it whn it rachs 0% of its original valu. Calculat th numbr of yars th machin will hav bn in us bfor bing rplacd. 8.6 Eponntial and Log Graphs A Eponntial Graphs. Eponntial Graphs from a Tabl of Valus and its Proprtis * Lik othr functions, w can construct an ponntial graph by forming a tabl of valus. * For ampl, considr th function f ( ) and th corrsponding graph y. * W can construct a tabl of valus for this function. 4 3 0 3 4 y 4 8 6 6 8 4 Pag 6
* From this tabl, w can construct th graph of th curv y. * Th ky faturs of this curv ar as follows. (i) Th curv lis compltly abov th -ais. (ii) On th lft, th curv gts closr and closr to th -ais. Th -ais is calld an asymptot of th curv. An asymptot of a curv is a lin which approimats th curv as th curv tnds to infinity. (iii) On th right, th curv incrass mor and mor rapidly as incrass.. Eponntial Growth Graphs * Th quation of an ponntial growth graph is of th form y a( b) whr a 0 and b. * Som of ths graphs ar shown opposit. Notic that: 3 4 0 8 6 4 y y 3 y 3( 5) y 4() (i) Th largr th bas, b, th mor stply th graph riss on th right. (ii) Th graph intrscts th -ais at th point ( a,0). 3. Eponntial Dcay Graphs * Th quation of an ponntial dcay graph is of th form y a( b) whr a 0 and 0 b. * Som of ths graphs ar shown opposit. Notic that: (i) Th smallr th bas, b, th mor rapidly th graph rducs on th right. (ii) Th graph intracts th -ais at th point ( a,0). y (05) y (07) y 05(3) y (05) Pag 7
B Logarithmic Graphs. Eponntial and Log Functions ar Invrss * Th functions y a and y log a ar invrs functions of ach othr. * To dmonstrat this, w show that prforming on aftr th othr rturns us to. (i) log a log a () a log (ii) To find a a, lt log a y a Thn loga y log y Thus log a a a. a. Graph of Log as Invrs of Eponntial * As discussd abov, th functions f ( ) and g( ) log ar invrss of ach othr. * Bcaus of this, th graph of ach is th rflction of th othr in th lin y. y y y log * Th two graphs, y and y log, ar shown opposit. 3. Othr Log Graphs * Th diagram opposit shows th thr log graphs: y log 3, y log 4 and y log0. * Each of ths graphs (i) is only dfind for 0, (ii) crosss th -ais at (,0), (iii) tnds to as tnds to 0 from th plus sid (th y-ais is an asymptot) (iv) has a y co-ordinat of whn th co-ordinat is qual to th bas. 0 y y log 3 y log 4 y log 0 4 6 8 0 4 6 8 Pag 8
4. Constructing Log Graphs W can construct a log graph by (i) making a tabl of valus, and (ii) using our knowldg of th shap of a log graph, as outlind abov. C Solving Equations Using Intrscting Graphs W can obtain approimat solutions of an quation containing ponntial functions or log functions by finding th intrsction of th typs of graphs w hav discussd prviously. Eampl (i) Sktch a rough graph of th function f : ( 5). (ii) Using th sam as and th sam scals, sktch a rough graph of th function g :. (iii) Us your graph to stimat th solutions of th quation ( 5). (iv) What othr mthods, if any, could hav bn usd to solv this quation? Solution (i) Th function is f ( ) ( 5) and th corrsponding curv is y ( 5). Constructing a tabl of valus: y 0 33 3 45 6 y ( 5) Th graph is shown blow. y 4 (ii) Th function is g( ) and th corrsponding curv is y. This curv is also shown on th graph opposit. 3 (iii) Th quation ( 5) may b writtn f ( ) g( ). Pag 9
Th solutions of this quation ar th co-ordinats of th points of intrsction of th curvs y f ( ) and y g( ). From th graph, th solutions ar and, corrct to on dcimal plac. (iv) W hav no algbraic mthod on our cours to solv th quation ( 5). On othr possibl mthod is to stimat th solutions from tabls of valus, but th graphical mthod is undoubtdly th bst availabl. Erciss 8.6. f : : 3(). (i) By complting th following tabl of valus, construct a graph of th curv y f ( ), for 3. y f ( ) 3 0. (ii) Using th sam as and th sam scals, construct a graph of th function g : 3, for 3. (iii) Us your graphs to find th solutions of th quation 3() 3. (iv) What othr mthods that w hav sn can b usd to solv this quation? Discuss. 3 f : :. 3 (i) By complting th following tabl of valus, construct a graph of th curv y f ( ), for 4. y f ( ) 0 3 4 (ii) Using th sam as and th sam scals, construct a graph of th function g :, for 4. (iii) Us your graphs to find th solutions of th quation 3. 3 (iv) What othr mthods that w hav sn can b usd to solv this quation? Pag 0
3. Two functions ar f : : (5) and g : : 3(0 5). (i) By complting th following tabl of valus, construct a graph of th curv y f ( ), for. y f ( ) 0 (ii) Using th sam as and th sam scals, sktch a graph of y g( ), for. (iii) Us your graphs to find th solution of th quation f ( ) g( ). (iv) Chck your answr to part (iii) by using algbra. 4. f : : log5. (i) Copy and complt th tabl: y f ( ) 5 5 5 5 Hnc construct th graph y f ( ). (ii) Using th sam as and th sam scals, sktch th graph of g :. (iii) Us your graph to find th solutions of th quation log5. (iv) What othr mthods that w hav sn can b usd to solv this quation? Discuss. 8.7 Natural Eponntial and Natural Log. Th Numbr * Th numbr is a famous and important mathmatical constant. Its valu is approimatly 78. * Th origin of this numbr will b sn in a latr chaptr.. Natural Eponntial Function * Th natural ponntial function, or somtims simply th ponntial function, is f ( ), whr is approimatly 78. Pag
* Th natural ponntial function obys all th usual Laws of Indics,.g. 3 3 (i). (ii) ( ). 3. Natural Log Function * Th natural log function is f ( ) log whr is approimatly 78. * log can also b writtn ln. Thus w can also writ f ( ) ln. * Th natural log function obys all th usual Laws of Logs,.g. (i) ln(. y) ln ln y (ii) ln 4 4ln. 4. Natural Eponntial and Natural Log ar Invrss Lik othr ponntial and log functions with th sam bas, th natural ponntial and natural log functions ar invrss. Thus (i) ln (ii) ln. y y 5. Graphs * As invrs functions, th graphs y and y ln ar rflctions of ach othr in th lin y. 0 y log ln * Ths graphs ar shown opposit. 6. Othr Eponntial and Log Graphs * Lik any ponntial or log graphs, w can plot a natural ponntial graph,.g. y 3, or a natural log graph,.g. y ln( 3), by making a tabl of valus. * Not that y riss mor rapidly than dcras mor rapidly than y y, for 0., for 0, and that y will Pag
Eampl Show that ln ln. Solution ln ln ln ln ln ln. 7. Eponntial Growth In scintific work, it is vry common to s a quantity, Q( t ) which grows ponntially with tim, t, bing prssd in th form bt Q( t) A, whr b is a positiv constant. 8. Eponntial Dcay Also in scintific work, it is vry common to s a quantity, Q( t ), which dcays ponntially with tim, t, bing prssd in th form Q( t) A bt, whr b is a positiv constant. Eampl In a laboratory primnt, a quantity, Q( t ), of a chmical was obsrvd at various points in tim, t. Tim is masurd in minuts from th initial obsrvation. Th tabl blow givs th rsults. t 0 3 4 Q( t) 59000 48305 39549 3 380 650 Q follows a rul of th form Q( t) A b t, whr A and b ar constants. (i) Us two of th obsrvations from your tabl to valuat A and b. Vrify your valus by taking anothr obsrvation from th tabl. (ii) Hnc find th valu of Q aftr 0 minuts, i.. Q (0). (iii) Estimat th tim that it taks for th quantity of th chmical to rduc to 0% of th original amount. Giv your answr in minuts, corrct to thr dcimal placs. t (iv) It is suggstd that anothr modl for Q is Q( t) Ap. Eprss p in trms of b, corrct to fiv dcimal placs. Pag 3
Solution (i) Q(0) 5 9 : (ii) b A (0) 59 0 A 59 A 59 () Q() 4 8305 : 59 b 48305 b 4 8305 59 b 088788 b ln 088788 b 0 b 0 0 t Thus Q( t) 59 To chck: 0(4) 08 Q(4) 59 59 650 Corrct. 0 (0) Q(0) 59 59 07895 (iii) 0% of th original amount is 0 59. Thus Q( t) 0 59 (iv) 0 59 t 059 0 t 0 0 t ln 0 0 t 30585 t 53 minuts. 0 t Q( t) 59 Q( t) 59 t 0 Q( t) 59(08873) t Thus p 0 8873, corrct to fiv dcimal placs. Erciss 8.7 Simplify ach of th following prssions.. log 4.. ln( ) 5. log sin 4 cos 3. log 3 sin ln( ) 6. ln 4 Pag 4
7. A chmical raction starts with 000 grams of ractant X, which thn rducs ponntially as th raction procds. Th amount of X, in grams, prsnt aftr t minuts is givn by A( t) C b t, whr C and b ar constants. Th following tabl givs som valus of A. t 0 3 4 A( t) 000 36788 (i) Find th valu of C and th valu of b. Hnc complt th tabl abov. (ii) Find th valu of t for which A( t) 50. (iii) Anothr ractant, Y, also starts with 000 grams and rducs ponntially with tim, so that th amount of Y rmaining, B( t ), is givn by B( t) C dt. If aftr 4 minuts, thr is lss of Y rmaining than of X, suggst a possibl valu for th constant d. 8. Th amount a company ows on a loan grows ponntially. Th amount owd aftr t yars, A( t ), is givn by th following tabl. t 0 3 4 A( t) 57446 595808 bt A is givn by a rul of th form A( t) P, whr P and b ar constants. (i) Find th valu of P and th valu of b, and hnc complt th tabl. (ii) Find th numbr of yars for th amount owd to doubl from its initial amount. (iii) If A( t ) can also b writtn A( t) P( i) t, find th valu of th constant i. 9. A bactria colony grows by 4% pr day. (i) If th siz at tim t 0 is 00, prss Q( t ), th siz aftr t days, in th form bt Q( t) A, whr A and b ar constants. (ii) Find, corrct to two dcimal placs, th tim it taks for th colony to doubl in siz. Q( t k) (iii) Show that is indpndnt of t. Q( t) 0. In 000, 00 grams of radium, a radioactiv substanc, wr stord. This quantity dcays ponntially ovr tim. Lt Q( t) A b t rprsnt th amount of radium rmaining t yars latr. (i) Th half-lif of radium is 60 yars, i.. it taks 60 yars for half of th original amount to dcay. Find th valus of th constants A and b. (ii) Find th mass of radium that will rmain in th yar 4000, corrct to two dcimal placs. (iii) Find th mass of radium that will dcay btwn th yars 3000 and 4000. Pag 5
Solutions to Erciss Erciss 8.5. (i) (ii). (i) (ii) 05 05 5765 55065 05, 05, 05, 05 00 05 05 5765 As ths ratios ar constant, y is an ponntial function of. y a( b) 0 0, y 00 : 00 a( b) a 00, y 05 : 05 00( b) b 05 Thus y 00( 05). 480 4608 44368 446738 096, 096, 096, 096 500 480 4608 44368 As ths ratios ar constant, y is an ponntial function of. y a( b) 0, y 500 :, y 480 : 0 500 a( b) a 500 480 500( b) b 096 b 09798 Thus y 500(0 9798). 3. (i) Calculating ratios of succssiv trms: 9000 800 790 656 09, 09, 09, 09 0000 9000 800 790 As th ratio is constant, W is an ponntial function of. (ii) Lt W a( b). Thn b 0 9, and 0 0000 a(09) a 0000 Thus Pag 6
W 0000(09) (iii) Whn 75, 75 W 0000(09) W 7484 57 g (iv) Half th original amount is 5000 g. 5000 0000(09) (09) 05 log0 9 05 6579 (v) W 000 : 000 0000(09) (09) 0 log0 9 0 854. 4. (i) f (0) 500 f () f (0) 4% of f (0) f () 96% of f (0) f () f (0) 096 500096 (ii) f () f () 096 (iii) 00 f () f (0) 096 096 f (0) (096) f (3) f (0) (096) f ( ) f (0) (096) 500(096) 500(096) 096 0 4 log0 96 0 4, corrct to th narst yar. 5. (i) f (0) N f () N 3% of N f () N( 03) f () N( 03) f ( t) N( 03) t (ii) N N( 03) t t 03 t log 03 t 3, corrct to th narst hour 6 (iii) 4386 N(03) 4386 N 6 03 N 000. 6. (i) A( t) P( 005) t A( t) P( 05) t (ii) A( t) 5P 3 Pag 7
P t ( 05) 5 t ( 05) 5 t log 5 05 P t 6 4 (iii) A( t) A(384) P(05) 5P P (iv) A(64 t) P 64t P(05) P 64t 05 05 3 84 64 t log t 65. 7. (i) f ( t) 00000( 008) t f ( t) 00000(09) t (ii) f ( t) 50000 t 00000(09) 50000 t (09) 05 t log0 9 05 t 8 3, corrct to two dcimal placs (iii) f ( t) 0000 t 00000(09) 0000 t (09) 0 t log0 9 0 t 9 3 yars. Erciss 8.6. (i) f ( ) 3() Tabl: 3 0 3 3 3 y f ( ) 3 6 8 4 Th graph is constructd blow. (ii) g( ) 3 Th linar graph y 3 contains th points ( 3, 0) and (0,3), and is shown opposit. Pag 8
(iii) From th graph, th solutions ar approimatly 5 and 0. (iv) W could compar valus in th tabls of valus, but thr is no algbraic mthod w hav sn that can b usd to solv this quation. 3. (i) f ( ) 3 Tabl: y f Th graph is constructd blow. (ii) g( ) Th linar graph y contains th points (0,) and (,), and is shown opposit. (iii) From th graph, th solutions of th quation ar approimatly 0 3 4 ( ) 03 0 44 067 5 5 3375 and 3 4. (iv) W could compar valus in th tabls of valus, but thr 3 is no algbraic mthod w hav sn that can b usd to solv this quation. 3 3. (i) f ( ) (5) Tabl: 0 y f ( ) 03 08 5 5 Th graph is constructd blow. (ii) g( ) 3(0 5) Tabl: y 3() 3 y y 4 3 y 3 y 3 y 3 Pag 9
0 y g( ) 6 3 5 075 Th graph is constructd blow. (iii) From th graph, th only solution of th quation is 0 5. (iv) f ( ) g( ) ( 5) 3(0 5) 5 3 y 3(0 5) 5 3 5 3 3 5 5 log55 059 4. (i) f ( ) log5 Tabl: 5 5 5 5 y f ( ) 0 Th graph is constructd blow. (ii) g( ) y Th linar graph y contains th points (, 0) y log 5 and (0, ), and is 3 shown opposit. (iii) From th graph, y th solutions of th quation ar 3 approimatly 0 and 7. (iv) W could compar valus in th tabls of valus, but thr is no algbraic mthod w hav sn that can b usd to solv this quation. 3 y y (5) Pag 0
Erciss 8.7.. 3. 4. 5. 6. log ( ) log log log ( )log log sin log (sin ) log sin log ( ) log log 4 cos 4 cos 4log (cos )log 4log cos ln( ) ln ln( ) ln ln( ) 3 sin 3 sin ln( ) ln ln 3ln sin (ln ) 3ln sin ln ln ln 4 4 ln (4 ) ln ln 4 7. (i) A( t) C bt A(0) 000 : A(4) 367 88 : Thus and (ii) A( t) 50 0 000 C C 000 367 88 000 b(4) 4b 036788 4b ln 036788 4b 09999 b 05 0 5t A( t) 000 A A A 0 5 () 000 7788 0 5 () 000 60653 0 75 (3) 000 4737 0 5t 000 50 0 5t 005 05t ln 005 05t 9957 t 98 (iii) d 0 5,.g. d 0 5. Pag
8. (i) A( t) P bt b A() 574 46 : 57464 P A() 5958 08 : b 595808 P Dividing by, b P 595808 b P 57464 b 040778 b ln040778 b 003997 : 57464 P(040778) P 550035 Thus 0 03997t A( t) 550035 and A(0) P 550035 A A (ii) A( t) P 3b (3) P 6004 4b (4) P 64539 0 03997t P P 0 03997t 003997t ln 003997t 069347806 t 734 (iii) A( t) P( i) t ( i) t bt b i i 040778 i 0040778 9. (i) Q( t) 00( 04) t b Lt 04 b ln04 00390735 0 0390735t Thus Q( t) 00 (ii) Q( t) 400 t 00(04) 400 t (04) t log 7 67 04 tk Q( t k) 00( 04) (iii) 04 t Q( t) 00( 04) which is indpndnt of t. 0. (i) Q( t) A b t Q(60) A A A 60b 05 60b k Pag
60b ln 05 60b 069347806 b 000043676 and A 00. (ii) In 4000, t 000. 0 00043676(000) Q(000) 00 Q(000) 4 09 grams (iii) In 3000, t 000. 0 00043676(000) Q(000) 00 Q(000) 6488 Th mass of radium that dcrass btwn 3000 and 4000 is 6488 409 79 grams. Pag 3