Basic Maths & Logarithm Solved Examples

SOLVED EXAMPLES Ex. 1 Sol. Find the value of x and y for which (2 + 3i) x2 – (3 – 2i) y = 2x – 3y + 5i where x, y  R. (2 + 3i)x2 – (3 – 2i)y = 2x – 3y + 5i  2x2 – 3y = 2x – 3y  x2 – x = 0  x = 0, 1  if x = 0, y =  x = 0, y = 5 2 5 2 and 3x2 + 2y = 5 and if x = 1, y = 1 and x = 1, y = 1  5 are two solutions of the given equation which can also be represented as  0,  & (1, 1)  2 Ex. 2 Sol.  5  0,  , (1, 1)  2 Show that (x – 3) is a factor of the polynomial x3 – 3x2 + 4x – 12. Let p(x) = x3 – 3x2 + 4x – 12 be the given polynomial. By factor theorem, (x – a) is a factor of a polynomial p(x) iff p(A) = 0. Therefore, in order to prove that x – 3 is a factor of p(x), it is sufficient to show that p(3) = 0. Now, p(x) = x3 – 3x2 + 4x – 12  p(3) = 33 – 3 × 32 + 4 × 3 – 12 = 27 – 27 + 12 – 12 = 0 Hence, (x – 3) is a factor of p(x) = x3 – 3x2 + 4x – 12. Ex. 3 Sol. If a two-digit number is divided by the number having same digits written in reverse order, we get 4 as quotient and 3 as remainder and if the number is divided by the sum of the digits then 8 as a quotient and 7 as a remainder is obtained. Find the number. Let 10x+y be the required number.  10x+y = 4(10y+x) + 3 ...... (i) and 10x+y = 8(x+y) + 7, ...... (ii) on solving (i) and (ii) we get x=7, y=1  the number is equal to 71 2 Ex. 4 1 1  If  a   = 3, then find value of a3 + 3 . a a   Sol. a+ 1 =± a 3 3 a3 + 1 1  1  a   – 3  a   = ± 3 3  3 3 = 0. 3 =  a a   a 4 4 Ex. 5 3 6 9  6 3 9  Simplify  a   a  .     Sol. a9(1/6)(1/3)4 . a9(1/3)(1/6)4 = a2 . a2 = a4 . BASIC MATHS & LOGARITHM Ex. 6 Sol. 43 5 =a+ b 5 43 5 Find rational numbers a and b, such that 43 5 43 5 × =a+ b 5 43 5 43 5 61 24 a=– ,b=– 29 29 61  24 5 =a+b 5 29  Ex. 7 What term must be added to each term of the ratio 5 : 37 to make it equal to 1 : 3? Sol. Let x be added to each term of the ratio 5 : 37. Then x5 1 = x  37 3  3x + 15 = x + 37 i.e. x = 11 3x 4  x 2  2x  3 5x 4  2x 2  7x  3 = 3x 4  x 2  2x  3 5x 4  2x 2  7x  3 Ex. 8 Solve the equation Sol. 3x 4  x 2  2x  3 5x 4  2x 2  7x  3 = 3x 4  x 2  2x  3 5x 4  2x 2  7x  3 By the process of componendo and dividendo, we have 3x 4 5x 4 = x 2  2x  3 2x 2  7x  3 or 3x4 (2x2 – 7x + 3) – 5x4(x2 – 2x – 3) = 0 or x4 [6x2 – 21x + 9 – 5x2 + 10x + 15] = 0 or x4 (x2 – 11x + 24) = 0  x=0 or x2 – 11x + 24 = 0 x=0 or (x – 8) (x – 3) = 0  Ex.9 x = 0, 8, 3 Solve the inequality if 5 1  (x – 2)10 (x  1)3  x –  (x  8) 2 2  f(x) = is > 0 or < 0. x 24 (x – 3)3 (x  2)5 Sol. Let 5 1  (x – 2)10 (x  1)3  x –  (x  8) 2 2  f(x) = x 24 (x – 3)3 (x  2)5 the poles and zeros are 0, 3, – 2, – 1, + + –8 + –2 – –1 1 , – 8, 2 2 + 0 + – 1/2 2 – 3   1 2 If f(x) > 0, then x (– , – 8) (– 8, – 2) (– 1, 0)   0,  (3, ) 1 2   and if f(x) < 0, then x (– 2, – 1)   , 2   (2, 3) Ex. 10 Solve the following linear equations (i) x |x| = 4 (ii) |x – 3| + 2|x + 1| = 4 Sol. (i) x|x| = 4 If x > 0  x2 = 4  x=±2  x=2 (  x  0) 2 2 If x<0  –x =4 x = – 4 which is not possible (ii) |x – 3| + 2|x + 1| = 4 case  : If x  –1  –(x – 3) – 2(x + 1) = 4  – x + 3 – 2x – 2 = 4  – 3x = 3 case  : If – 1 < x  3  –(x – 3) + 2(x + 1) = 4  – x + 3 + 2x + 2 = 4  x = – 1 which is not possible case  : If x > 3 x – 3 + 2(x + 1) = 4 3x – 1 = 4   x=–1 Ex. 11 5 = 2.236 and If   x = 5/3 – 3x + 1 = 4 x=–1 which is not possible 2 = 1.414, then evaluate : 3 4  5 2 5 2 3 4  5 2 5 2 Sol. 3 = =    5  2 2  5  2  5 2 4  5 3 5 3 2  4 5  4 2 52 7 5 2 7 5 2  = 52 3 15.652  1.414 7  2.236  1.414 = c = = 3 3 17.066  5.688 ((Approx). = 3 = Ex. 12 Sol. If 25x–1 = 52x–1 – 100, find the value of x. We have,  25x–1 = 52x–1 – 100  (52)x–1 = 52x–1 – 100.   52x–2 – 52x–2.51 = –100  2x–2  5 (–4) = –100   52x–2 = 52   2x = 4 x = 2. 52x–2 – 52x–1 = –100 52x–2 (1 – 5) = –100 52x–2 = 25 2x – 2 = 2 BASIC MATHS & LOGARITHM Ex. 13 Find the value of the followings : (i)  1  1  1 log a 1    log a  1    .......  log a  1   2 3      n (ii) log 2 75 5 32 – 2 log 2 + log 2 16 9 243 1 log 53 (iii) 81 Sol. (i)  1  1  1 log a 1    log a  1    .......  log a  1   2 3      n 1 2 2 3  n –1  1 1 2 3 n –1 = loga   = – logan  = loga  · · .......   n  n n  2 3 4 = loga   + loga   + ...... + loga  (ii)  75  5  32   – 2log2   + log2    16  9  243  log2   75   25   35   – log2   + log2    16   81   243  = log2   75 32 81     = log 2 = 1 2  16 243 25  = log2  (iii) 81 1 log 53 log 5 3 = 81 = 3 log 3 5 4 = 5 4 = 625 Ex. 14 log2 log4 log5 x = 0 Sol. log4log5x = 20 = 1 log5x = 4  x = 54 = 625 Ex. 15 The value of N, satisfying loga[1 + logb{1 + logc(1 + logpN)}] = 0. Sol. 1 + logb{1 + logc(1 + logpN)} = a0 = 1  logb{1 + logc(1 + logpN)} = 0  1 + logc(1 + logpN) = 1  logc(1 + logpN) = 0  1 + logpN = 1  logpN = 0  N = 1 Ex. 16 If Sol. y3  b3 ( x  y  z)3  (a  b  c)3 x y z x3  a3 z 3  c3 = = , then show that 2 + + = . (x  y  z) 2  (a  b  c)2 y2  b2 a b c x  a2 z2  c2 x y z = = = k (constant) a b c  x = ak; y = bk; z = ck Substituting these values of x, y, z in the given expression y3  b3 (x  y  z)3  (a  b  c)3 x3  a3 z 3  c3 + + = 2 2 y b (x  y  z) 2  (a  b  c)2 x2  a2 z2  c2 we obtain L.H.S. = a 3 (k 3  1) b3 (k 3  1) c3 (k 3  1) a 3k3  a 3 b3k 3  b3 c3k 3  c3 + + = + + a 2 (k 2  1) b 2 (k 3  1) c 2 (k 2  1) a 2k2  a 2 b2 k 2  b2 c2 k 2  c2 = (k 3  1) a(k 3  1) b(k 3  1) c(k 3  1) + + = . (a + b + c) (k 2  1) k2 1 k2 1 k2  1 (ak  bk  ck)3  (a  b  c)3 k 3 (a  b  c)3  (a  b  c)3 = = 2 (ak  bk  ck) 2  (a  b  c) 2 k (a  b  c) 2  (a  b  c)2 Now R.H.S (k 3  1)(a  b  c)3 (k 3  1) = = 2 . (a + b + c) (k 2  1)(a  b  c) 2 (k  1) We see that L.H.S. = R.H.S. Ex. 17 If a2 + b2 = 23ab, then prove that log Sol. 2 2 (a  b) 1  (log a  log b) . 5 2 2 a + b = (a + b) –2ab = 23ab 2  (a + b) = 25ab  a+b = 5 ab Using (i) L.H.S. = log ....(i) (a  b) 5 ab 1 1  log  log ab  (log a  log b) = R.H.S. 5 5 2 2 Ex. 18 If z 1 is purely imaginary, then prove that | z | = 1 z 1 Sol.  z 1   =0  z 1 Re   z  1  z 1  +  =0 z  1  z 1   z 1 z 1 + =0 z 1 z 1  zz – z +z–1+zz –z+ z –1 =0   zz =1 |z|=1 Ex. 19 Evaluate : 811 / Sol. 81 log 3 5 3 | z |2 = 1  Hence proved log 5 3 3 log 9 36 3  27 log9 36  3 4 / 4 log 9 7 3/ 2  3 4 log3 5  3 log3 (3 6 )  3 log3 7 = 625 + 216 + 49 = 890. 2 log7 9 BASIC MATHS & LOGARITHM Ex. 20 If a, b, c are distinct positive real numbers different from 1 such that (logba . logca – logaa) + (logab . logcb – logbb) + (logac . logbc – logcc ) = 0, then find value of abc. Sol. (logba logca – 1) + (logab . logcb – 1) + (logac logbc – 1) = 0  log a log a log b log b log c log c .  .  .  3  log b log c log a log c log a log b  (loga + logb + logc) = 0  log abc = log 1  [If a3 + b3 + c3 – 3abc = 0, then a + b + c = 0 if a  b  c] abc = 1 Ex. 21 If ||x–1| – 2| = 5, then find x. Sol. |x – 1| – 2 = ± 5 |x – 1| = 7, – 3 Case-I : When |x – 1| = 7  x – 1 = ± 7  Case-II : When |x – 1| = –3 (reject) Ex. 22 Sol. (log a)3 + (log b)3 + (log c)3 = 3loga logb logc x = 8, –6 Solve the inequality log1/3 (5x – 1) > 0. by using the basic property of logarithm.  5x – 1  1   5x – 1  0     2 2 x x   5x5x       5x5x  1 1 x x   2 5 1 5 1 2 The solution of the inequality is given by  ,  5 5 Ex. 23 Solve the inequality log(2x + 3) x2 < log(2x + 3) (2x + 3). Sol. The given inequality is equivalent to the collection of the systems  0  2x  3  1  2   x  2x  3  2x  3  1    0  x 2  2x  3 .........(i) .........(ii) Solving system (i) we obtain  –3   x  –1 2 (x – 3) (x  1)  0 .........(iii) Ans. System (iii) is equivalent to the collection of two systems  –3  2  x  –1, x  3   –3  x  –1, x  –1  2 .........(iv) .........(v)  –3  , – 1 ,  2  system (iv) has no solution. The solution of system (v) is x   solving system (ii) we obtain.  x  –1   (x – 3) (x  1)  0  x  –1   –1  x  3 or  x  (– 1, 3)  –3  , –1 (– 1, 3) 2   x  Ex. 24 Solve the inequation log  2   x –12 x  30    10   Sol. 2x    log 2  > 0. 5   This inequation is equivalent to the collection of following systems.  x 2 – 12x  30  1,   10  and  log  2x   1,  2  5  Solving the first system we have.  x 2 – 12x  30  1, 0   10   0  log  2x   1 2    5   x 2 – 12x  20  0   2x 2   5  (x – 10) (x – 2)  0  x  5   x  2 or x  10  x  5 Therefore the system has solution x > 10 Solving the second system we have.     0  x 2 – 12x  30  10   2x 2 1  5  and x2 – 12x + 20 < 0   x 2 –12x  30  0  5  x5 2 x  6 6  x  6 – 6  5  x5 2 or The system has solutions equation is. x( 5 ,6– 2 5 < x < 6 – 2 6 (10, ) and 2  x  10 6 combining both systems, then solution of the original in BASIC MATHS & LOGARITHM Ex. 25 If |x – 1| + |x + 1| = 2, then find x. Sol. Case-I If x  –1 –(x – 1) – (x + 1) = 2  –x + 1 – x – 1 = 2  –2x = 2  x = –1 ........(i) Case-II If – 1 < x < 1 –(x – 1) + (x + 1) = 2  –x+1+x+1=2  2 = 2  –1 < x < 1 ........(ii) Case-III If x  1 x–1+x+1=2  x=1 ........(iii) Thus from (i), (ii) and (iii) – 1  x  1 Ex. 26 Solve : x |x + 3| + 2 |x + 2| = 0 Sol. Case-I x<–3 –x(x + 3) – 2 (x + 2) = 0 x2 + 5x + 4 = 0  x = –1, –4  x = –4.      x = –1 (reject) Case- II –3 < x < – 2 (x) (x + 3) – 2x –4 = 0 x2 + x – 4 = 0  x 1  17 1  17 , 2 2  x= 1  17 2     x 1  17 (reject) 2 Case-III x > –2 x (x + 3) + 2x + 4 = 0 x2 + 5x + 4 = 0 Hence x = –4,  x = –1, – 4.  x = –1  x = –4 (reject)  1  17 , –1. 2 31 Ex. 27 Solve the equation [x] + {–x} = 2x, (where [.] and { . } represents greatest integer function and fractional part function respectively). Sol. Case- x x + 0 = 2x  x=0 Case- x [x] + 1 – {x} = 2x [ + f] + 1 – { + f} = 2 ( + f)  + 1 – f = 2 + 2f 1  = fas 3 0< 0 0 both numbers p and q may be regarded as natural number  3 = 2p/q  2p = 3q But this is not possible for any natural number p and q. The resulting contradiction completes the proof. Ex. 32 If in a right angled triangle, a and b are the lengths of sides and c is the length of hypotenuse and c – b  1, c + b  1, then show that logc+ba + logc–ba = 2logc+ba . logc–ba. Sol. We know that in a right angled triangle c2 = a 2 + b 2 c2 – b 2 = a 2 LHS = .....(i) 1 1 log a (c  b)  log a (c  b)   log a (c  b) log a (c  b) log a (c  b). log a (c  b)  log a (c2  b2 ) log a a 2  log a (c  b). log a (c  b) log a (c  b).log a (c  b)  2 log a (c  b). log a (c  b) = 2log(c+b)a . log(c – b) a = RHS (using (i)) 6 loga x.log10 a.loga 5 log10 ( x/10) log100 xlog4 2 a 3 9 5 Ex. 33 Solve the following equation for x : Sol. Let A = logax . log10a . loga5 = log10x . loga5 = log a (5 )log10 x  aA = 5 log10 x  5  (say log10 x   ) Let B = log10(x/10) = log10x – 1 =  – 1  3B = 3– 1 = Let 1 1 C = log100x + log42 = log1 0 2 x  log 2 2 2 =  9C  9 1 2 3 3 1 1  1 log10 x   2 2 2  3 1  3 .3  According to question 6  3 .5   3.3  5 3 1  6.5– 1 = 3   3   6.5– 1 = 3– 1(10)  5– 2 = 3– 2 3 which is possible only when  = 2  log10x = 2  x = 102 = 100  Ex. 34  x4  2  Solve : log(2x–1)   =1 2x 1 Sol. x4  2 = 2x – 1  x4 + 2 = 4x2 – 1 2x 1  x4 – 4x2 + 3 = 0  x   3,  1  x2 = 4 4 2 16  12 4 4  = = 3, 1 2 2 2 ......... (i)  x4  2  log  we get 2x – 1 negative. And if x = 1 in 2x – 1 2 x 1  Substituting x =  3 and –1 in  2x 1  we get base = 1  reject x = ±1,  3 Hence x  3