============================================================ W zadaniach korzystamy z definicji logarytmu: Jeżeli a > 0 i a jest różne od 1 i b > 0 , to rozwiązanie równania a^x = b nazywamy logarytmem liczby dodatniej b przy podstawie a. tzn. log a ( b ) = x <=> a^x = b ============================================================ z.1 a) log 2 (x) = - 3 <=> 2^(-3) = x <=> x = 1/2^3 = 1/8 x = 1/8 ======= b) log 3 ( 8 - x) = 0 <=> 3^0 = 8 - x <=> 8 - x = 1 <=> x = 7 x = 7 ======= c) log 3( x^2 + 2) = 3 <=> 3^3 = x^2 + 2 <=> x^2 + 2 = 27 <=> x^2 = 25 x = - 5 lub x = 5 ==================== d) log 1/2 (4x - 3) = - 3 <=> (1/2)^(-2) = 4x - 3 ,=> 2^2 = 4x - 3 <=> 4 = 4x - 3 <=> <=> 4x = 7 x = 7/4 ======== z.2 a) log 4 log 3 log 2(x) = 0 <=> 4^0 = log 3 log2 (x) <=> log 3 log 2(x) = 1 <=> <=> 3^1 = log 2 (x) <=> log 2 (x) = 3 <=> 2^3 = x <=> x = 8 x = 8 ===== b) log 4 log 2 log 3 (x -1) = 1/2 <=> 4^(1/2) = log2 log 3 ( x -1) <=> <=> 2 = log2 log 3 (x-1) <=> log 3(x-1) =2^2 = 4 <=> 3^4 = x -1 <=> <=> x -1 = 81 x = 82 ====== c) log (x -2) = log ( 5x - 4) <=> x - 2 = 5x - 4 <=> 4x = 2 x = 1/2 oraz x - 2 > 0 i 5x - 4 > 0 czyli x > 2 i x > 4/5 Brak rozwiązania =================== d) log 0,8 ( x^2) = log 0,8 ( 3x) x^2 = 3x i x^2 > 0 i 3x > 0 x^2 - 3x = 0 i x^2 > 0 i x > 0 x*( x - 3) = 0 i x > 0 x = 3 ======== z.3 a) log x 4 = 1/2 <=> x^(1/2) = 4 <=> x = 4^2 = 16 x = 16 ======= b) log x ( 1/81) = 4 <=> x^4 = 1/81 <=> x = 1/3 x = 1/3 ========= c) log x ( 2 - x) = 2 <=> x^2 = 2 - x <=> x^2 + x - 2 = 0 delta = 1 - 4*1*(-2) = 1 + 8 = 9 p(delty) = 3 x = [ - 1 - 3]/2 = -2 < 0 - odpada, bo x > 0 x = [ -1 + 3]/2 = 1 - odpada , bo musi być różne od 1 ================== d) log x ( 3 x^2 - 5x + 2) =2 <=> x^2 = 3 x^2 - 5 x + 2 <=> 2 x^2 - 5x + 2 = 0 delta = 25 - 4*2*2 = 25 - 16 = 9 p(delty) = 3 x = [5 - 3]/4 = 1/2 lub x = [ 5 + 3]/4 = 2 Oba spełniają założenia ======================= z.4 a) log 7 (x +3) - log 7 (x -1) = log 7 (5) założenia: x +3 > 0 i x - 1 > 0 czyli x > - 3 i x > 1 x > 1 ===== Mamy log 7 [ (x +3)/(x -1) = log 7 ( 5) (x +3)/(x -1) = 5 x +3 = 5*(x -1) x + 3 = 5 x - 5 4 x = 8 x = 2 =============== b) 2 log 5 (2 -x) - log 5 (x) - log 5 ( x + 5) = 0 Założenia: 2 - x > 0 i x > 0 i x + 5 > 0 czyli x < 2 i x > 0 i x > - 5 0 < x < 2 ========== Mamy log 5 (2 -x(^2 - [ log 5 (x) + log 5(x+5) ] = 0 log 5 ( 4 - 4x + x^2 ) - log 5 [ x *(x + 5)] = 0 log 5 [ ( x^ - 4 x + 4)/ ( x^2 + 5 x) ] = 0 = log 5 (1) ( x^2 - 4x + 4)/ ( x^2 + 5 x ) = 1 x^2 - 4 x + 4 = x^2 + 5 x 9 x = 4 x = 4/9 ======== c) log (2x + 5) + log ( x +3) = log (2x + 11) założenia: 2x + 5 > 0 i x +3 > 0 i 2x + 11 > 0 czyli x > - 2,5 i x > - 3 i x > - 5,5 x > - 2,5 ======= Mamy log [ ( 2x + 5)*(x + 3)] = log (2x + 11) ( 2x + 5)*(x + 3) = 2x + 11 2 x^2 + 6 x + 5x + 15 = 2x + 11 2 x^2 + 9 x + 4 = 0 ------------------------- delta = 9^2 - 4*2*4 = 81 - 32 = 49 p (delty) = 7 x = [ - 9 - 7]/4 = - 4 < -2,5 - odpada lub x = [ - 9 + 7]/4 = -2/4 = -1/2 x = - 1/2 =========
