XMS Logarithm practice

Question 1 Solve xxx 2log3log)1(log2 222 +=−− . Question 2 Solve 2)42(loglog450log 5255 ++=+ xx . Question 3 (a) Given that px=2log and qy=4log , express the following in terms of p and/or q. (i) x2log , (ii) 2 2log xy , (iii) y x4log 4 . (b) Solve the equation 02)]1(5lg[lg =−++ xx . Question 4 Solve the equation (a) 3log 2log 2 2 =+ xx . (b) 1)1(log)1(log 33 =−−+ xx . Question 5 Solve the equation )4(log3)2(log 33 −−=+ xx . Question 6 Given that )(logloglog 224 yxyx +=+ , express y in terms of x. Question 7 Solve the equation 5log)2(log2 1)12(log 25 2 33 =+−− xx . Logarithm

Solutions Question 1 (reject) 3 1or 5 1 0)13)(15( 01215 1612 16)1( )2(8log)1(log 2log2loglog)1(log 2log3log)1(log2 2 22 2 2 2 2 2 3 22 2 2 222 −= =+− =−+ =+− =− =− +=−− +=−− x xx xx xxx xx x xx x xxx xxx Question 2 (reject) 1or 2 0)1)(2( 02 1005050 )42(25log50log )42(25loglog250log 5log)42(log25log log450log 2)42(loglog450log 2 2 5 2 5 555 2 55 5 5 5 5255 −= =+− =−− += += +=+ ++=+ ++=+ x xx xx xx xx xx xx xx Question 3 (a) (i) pp x xx 0.5or 2 1 log2 1 loglog 2 2 1 22 = = =

Alternative Method, p xx p 2 1 2log2 1 loglog 2 2 1 22 = = = (ii) 2 log 4log loglog 2 2 2 4 y yy = = qy qy 2log log 2 1 2 2 = = qp qp yxxy 4 )2(2 log2loglog 22 2 2 += += += Alternative method, q qy 22 4 = = qp qp p yp yxxy q 4 )2(2 )2(log2 )(log2 logloglog 2 2 2 2 22 2 2 += += += += +=

(iii) 2 22or 2 2or 2 11 4log log1 loglog4log4log 2 2 4444 qp qp qp qx yx y x −+ −+ −+= −+= −+= Alternative method, qp qp qp y x qp q p q p −+= −+= −+= =      =      = −+ + 1 2 1 )22( 2 1 2log)22( )2(log 2 2log 4 )2(4log4log 4 22 4 2 2 4 44 (b) 4or (rej) 5 0)4)(5( 020 010055 10)55( 2)]1(5[lg 02)]1(5lg[lg 2 2 2 −= =−+ =−+ =−+ =+ =+ =−++ x xx xx xx xx xx xx

Question 4 (a) 2 4 1logor 2log 0)1)(log2(log 02log3)(log 3 log 2log 22 22 2 2 2 2 2 == == =−− =+− =+ xx xx xx xx x x Or Let y = log2 x 2 4 1log 2log 1 or 2 0)1)(2( 023 22 2 == == == =−− =+− xx xx yy yy yy (b) 2or (reject) 3 1 0)2)(13( 0273 01)12(3 0)1()1(3 3)1( 1 1)1( 1log 1)1(log)1(log 1)1(log2)1(log 1 3log2 1 )1(log)1(log 1 3log )1(log)1(log 1)1(log)1(log 2 2 2 1 2 23 2 33 33 3 3 3 2 1 3 3 3 33 = =−− =+− =−−+− =+−− =− + =− + =−−+ =−−+ =−−+ =−−+ =−−+ x xx xx xxx xx x x x x xx xx xx xx xx

Question 5 7or (NA) 5 0)7)(5( 0352 382 3)4)(2(log )4(log3)2(log 2 32 3 33 −= =−+ =−− =−− =−+ −−=+ x xx xx xx xx xx Question 6 ( ) 1or 1 1 )(loglog )(logloglog )(loglog2log log )(logloglog 2 1 2 1 2 2 1 2 22 2 1 2 222 2 2 224 − + − = =− =− += +=      +=+ +=+ +=+ x xxx x xy xxy xyyx yxyx yxyx yxyx yxyx yxyx

Question 7 (rejected) 1or 5 0)5)(1( 054 63144 3)2( )12( 1)2( )12(log 1)2(log)12(log 1)2(log)12(log2 2 1)2(log2 1)12(log 5log)2(log2 1)12(log 2 22 2 2 2 2 3 2 3 2 3 2 33 2 3