geiring.de > Math > Math 11 > Lösung Buch S. 71, 11 a)   `## 3^x = 5 | ln(::) ln(3^x) = ln(5) x*ln(3) = ln(5) x_1 = (ln(5))/(ln(3)) x_1 ~~ 1,465 ##` b)   `## 2,5^x = 7 | ln(::) ln(2,5^x) = ln(7) x*ln(2,5) = ln(7) x_1 = (ln(7))/(ln(2.5)) x_1 ~~ 2,12 ##` c)   `## 3*5^(x-2) = 7,2 | :3 5^(x-2) = (7,2)/3 | ln(::) (x-2)*ln(5) = ln((7,2)/3) (x-2) = ln((7,2)/3)/ln(5) x_1 = ln((7,2)/3)/ln(5)+2 x_1 ~~ 2,54 ##` d)   `## 0,5^x-2,5 = 0,5^(x+2) 0,5^x-2,5 = 0,5^x*0,5^2 | -rS. 0,5^x-0,5^x*0,5^2-2,5 = 0 | +2,5 0,5^x(1-0,5^2) = 2,5 0,5^x(1-1/4) = 2,5 0,5^x*3/4 = 2,5 | *4/3 0,5^x = 2,5*4/3 0,5^x = 10/3 | ln(::) x*ln(0,5) = ln(10/3) x_1 = (ln(10/3))/ln(0,5) x_1 ~~ -1,74 ##`