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Solving Exponential Equations. Example1. Solve. 3 2x+1 = 81. 3 2x+1 = 3 4. 2x + 1 = 4. 2x = 3. x = 3/2. The method in example one works well when the bases can be written as the same but consider the next equation. Example2. Solve. 6 x = 42. Logs must be used to solve this!.
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Solving Exponential Equations
Example1. Solve. 32x+1 = 81 32x+1 = 34 2x + 1 = 4 2x = 3 x = 3/2
The method in example one works well when the bases can be written as the same but consider the next equation.
Example2. Solve. 6x = 42 Logs must be used to solve this! log 6x = log 42 x log 6 = log 42 x = (log 42)/(log 6) x ≈ 2.086
5log8 + log5 2log 8 - log5 Example3. Solve. 82x-5 = 5x+1 log 82x-5 = log 5x+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 = ≈ 4.7095
log 25 log 4 Example4. Solve. y = log425 4y = 25 log 4y = log 25 ylog 4 = log 25 y = ≈ 2.322
Loga Log b The last example illustrates the change of base formula. The change of base formula states: Logba =
Example5. Solve. -3 log 2 2 log 2 - 3 log 3 22x+3 = 33x log 22x+3 = log 33x (2x+3) log 2 = 3xlog 3 2xlog 2 + 3log 2 = 3xlog 3 2xlog 2 - 3xlog 3 = -3log 2 x(2log 2 - 3log 3) = -3log 2 ≈ 1.089 t =
Example6. Solve. 2n = √ 3n-2 2n = (3n-2)1/2 2n = 3((n/2)-1) log 2n = log 3((n/2)-1) nlog 2 = ((n/2)-1)log 3 nlog 2 = (n/2)log 3 - log 3
Example6. Solve. -log 3 log 2 - (1/2) log 3 log 2n = log 3((n/2)-1) nlog 2 = ((n/2)-1)log 3 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 t = ≈ -7.638