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Exponential and Logarithmic Equations Exponential Equations Exponential equations are those equations where your variable x appears in the exponent. For example 2x = 10. To solve such equations for x; that is to find the value for x that makes the equation true, we need to bring the exponent down (some how!) and transform the equation to a polynomial equation. To do so, we use logarithmic functions. We use the fact log function is one-to-one; that is, if two numbers are equal, their logarithm is also equal (with any base a, but we can make wise choices to make our calculations easier): 2x = 10 ⇒ log(2x ) = log 10. Then we use the Laws of logarithms to bring the exponent down. Recall that loga B C = C loga B. Applying this law to our example we get log(2x ) = log 10 ⇒ x log 2 = log 10 Have in mind that log 2 and log 10 are just some real numbers, so treat them like numbers and solve the equation for x; x log 2 = log 10 ⇒ x = log 10 log 2 At this point it usually comes in handy of we know how two logarithmic function with different basis are related. Change of Basis Formula: The following formula, known as the change of basis formula, is very useful logb x = loga x loga b Example: Suppose your calculator only has a button for log10 . Use the change of basis formula to calculate log2 3. Examples: Solve the following equations. 1 3x−4 = 5 3e12x = 25 4(1 + 105x ) = 9 e3−x = 4 Here are the steps you take when you have only one exponential term with the variable x in the exponent. (like the above examples): 1. Isolate the only term with x in the power. So your equation looks like (number)expression with x = number 2 2. Take log from both sides. 3. Use the Laws of logarithm to bring the expression with x down: (expression with x) (log(number)) = log(number) 3 To solve the exponential equations where there are more that one exponentials, try one of the following: 1. If possible, use the laws of exponentials to reduce the problem to the case with only one exponential term with x in the exponent. For example: 23x 2x−5 = 5 2. If the equation looks like a quadratic equation, take everything to one side and set z = ax and solve the quadratic equation for z. For example 32x − (3x ) = 6 3. If the equation has no constant term, take everything to one side and factor the exponential. For example ex x2 = ex 4x3 e−3x − 3x4 e−3x = 0 4 Logarithmic Equations Logarithmic equations are those where your variable x appears in the argument of a logarithmic function. For example log3 (x + 5) = 12 Here are the steps you take to solve the logarithmic equations with only one logarithmic term with x appearing in the logarithmic argument: 1. Isolate the logarithmic term. 2. Transform the equation into exponential form. 3. Solve for x. Example:Solve the following equations. log2 (x − 2) = 4 4 + 3 log(2x) = 16 If you have more than one logarithmic term with x, you need to use the laws of logarithm to combine the log functions. 5 Example: Solve the following equations. log(x + 2) + log(x − 1) = 1 log5 (x + 1) − log5 (x − 1) = 2 Important: We only accept those solutions that are in the domain of our functions. log(x + 3) = log x + log 3 Example An amount of $6500 is invested in an account that pay 6% interest per year, compounded continuously. a) What is the amount after two years? b) How long will it be for the amount to be $8000? 6