Natural Logarithms Learning Objective Solve ln and ex

Natural Logarithms Learning Objective: §Solve ln and ex equations

The laws of natural logs Laws of logs: § ln a + ln b = ln ab § ln a - ln b = ln (a/b) § § ln a k = k ln a ln e = 1 ln ex = x ln e = x eln x = x

Example 1 Show e -ln 4 = 1/4 ln e -ln 4 = ln (1/4) -ln 4 ln e = ln (1/4) -ln 4 = ln (1/4) -1 x ln 4 = ln (1/4) ln 4 -1 = ln (1/4) QED

Example 2 § Solve: ln (2 x -3) = 1. 5 eln (2 x -3) = e 1. 5 2 x -3 = e 1. 5 2 x = 3 + e 1. 5 x = 1/2(3 + e 1. 5) eln x = x

Example 3 § Simplify: 1/ 2 ln 16 - ln 2 = ln 161/2 - ln 2 ln a k = k ln a = ln 16 - ln 2 = ln 4 - ln 2 =ln 4/2 = ln 2 ln a - ln b = ln (a/b)

Example 4 § Solve: e 2 x- 6 ex = 7 Set y = x e So, e 2 x = y 2 The equation becomes. . . y 2 - 6 y = 7 y 2 - 6 y - 7 = 0 (y -7)(y + 1) = 0 Either: (y-7)=0 y=7 or: (y+1)=0 y=-1 ex = 7 ex = -1 x = ln 7 x = ln -1

§ Solve: Example 5 5 - ex = 4 e-x Set y = ex The equation becomes. . . 5 - y = 4/y 5 y - y 2 = 4 y 2 - 5 y + 4 = 0 (y -4)(y - 1) = 0 Either: (y-4)=0 y=4 or: (y-1)=0 y=1 So, e-x = 1/y ex = 4 ex = 1 x = ln 4 x = ln 1

A bit of a summary When you have single exponentials When you have logs Log both sides e 2 x = e 6 ln e 2 x = ln e 6 2 x ln e = 6 ln e 2 ln x = 8 ln x 2 = 8 e ln x’ = e 8 x 2 = e 8 When you have different exponentials Find the exponent of both sides You may need to manipulate first e 2 x +4 ex +3 = 0 Use substitution y 2 +4 y +3 = 0 y = ex x ++3)( 3)(y (e(y ex + 1) = 0