log x + log( x -1) = log 20; założenie x > 1 log[ x*(x -1)] = log 20 x*(x -1) = 20 = 5*4 x = 5 ==== log 1/3 ( x + 10) + log 1/3 ( 7 - 2x) = - 4 założenia: x + 10 > 0 i 7 - 2x > 0 x > - 10 i x < 3,5 zatem log 1/3 [ (x + 10)*( 7 - 2x)] = log 1/3 ( 81) ( x + 10)*( 7 - 2x) = 81 7x - 2 x^2 + 70 - 20 x = 81 2 x^2 + 13 x + 11 = 0 delta = 13^2 - 4*2*11 = 169 - 88 = 72 = 36*2 p( delty) = 6 p(2) x = [ - 13 - 6p(2)]/4 lub x = [ - 13 + 6 p(2)]/4 ============================================= log 2 I x + 1 I = - 1 <=> I x + 1 I = 2^(-1) = 1/2 czyli x + 1 = - 1/2 lub x + 1 = 1/2 x = - 1,5 lub x = - 0,5 ======================= 2^( x - 1) = 4 2^(x -1) = 2^2 x - 1 = 2 x = 3 ======= 4^( x + 3) - 4^x = 504 4^3 * 4^x - 4^x = 504 64 * 4^x - 4 ^x = 504 63 *4^x = 504 / : 63 4^x = 8 (2^2)^x = 2^3 2^(2x) = 2^3 2x = 3 x = 1,5 ========
