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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.
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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
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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)
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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
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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.
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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?
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