1. Solving Exponential Equations

2. Example1. Solve. 3 2x+1 = 81 3 2x+1 = 3 4 2x + 1 = 4 2x = 3 x = 3/2

3. The method in example one works well when the bases can be written as the same but consider the next equation.

4. Example2. Solve. 6 x = 42 Logs must be used to solve this! log 6 x = log 42 x log 6 = log 42 x = (log 42)/(log 6) x ≈ 2.086

5. Example3. Solve.8 2x-5 = 5 x+1 log 8 2x-5 = log 5 x+1 (2x-5)log 8 = (x+1)log 5 2xlog 8 - 5log8 = xlog 5 +1log5 2xlog 8 - xlog5 = 5log8 +log5 x(2log 8 - log5) = 5log8 +log5 x = 5log8 + log5 2log 8 - log5 ≈ 4.7095

6. Example4. Solve. y = log 4 25 log 4 y = log 25 ylog 4 = log 25 y = log 25 log 4 4 y = 25 ≈ 2.322

7. The last example illustrates the change of base formula. The change of base formula states: Log b a = Log a Log b

8. Example5. Solve. 2 2x+3 = 3 3x log 2 2x+3 = log 3 3x (2x+3) log 2 = 3xlog 3 t = -3 log 2 2 log 2 - 3 log 3 2xlog 2 + 3log 2 = 3xlog 3 2xlog 2 - 3xlog 3 = -3log 2 x(2log 2 - 3log 3) = -3log 2 ≈ 1.089

9. Example6. Solve. 2 n = (3 n-2 ) 1/2 nlog 2 = ((n/2)-1)log 3 2 n = √ 3 n-2 2 n = 3 ((n/2)-1) log 2 n = log 3 ((n/2)-1) nlog 2 = (n/2)log 3 - log 3

10. Example6. Solve. t = -log 3 log 2 - (1/2) log 3 nlog 2 = ((n/2)-1)log 3 log 2 n = log 3 ((n/2)-1) nlog 2 = (n/2)log 3 - log 3 nlog 2 - (n/2)log 3 = -log 3 n(log 2 - (1/2)log 3) = -log 3 ≈ -7.638