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Here are a few recommended readings before getting started with this lesson.
An exponential equation is an equation where variable expressions occur as exponents. As with any kind of equation, there are different types of exponential equations.
Example Equation | |
---|---|
With One Variable | 2x=32 |
With the Same Variable on Both Sides | 22x=5⋅2x |
With the Same Base | 43x=42x+3 |
With Unlike Bases | 3x+4=81x |
With a Rational Base | (21)x=8 |
Emily and her friends went to the beach on a cloudy afternoon and cooked some chapati.
Since they did not smoke the chapati and did not use salt in the cooking process, Emily knows that it will not be too long before the food spoils. Food spoilage is generally caused due to the action of microorganisms like fungi. Fungal growth is exponential and given by the following exponential function.logca=logbclogba
Use a calculator
LHS⋅2=RHS⋅2
Rearrange equation
Round to nearest integer
t≈20
b1⋅a=ba
Calculate quotient
(ba)m=bmam
a⋅cb=ca⋅b
Use a calculator
An exponential inequality is an inequality that involves exponential expressions.
Let b be a positive real number different than 1. The following statements hold true.
The statements will be proved one at a time.
If b is greater than 1, the exponential function f(x)=bx is increasing for its entire domain.
In the graph, it can be seen that f(x1)≥f(x2) if and only if x1≥x2. Considering the definition of f(x), the statement is proven.If b is greater than 0 and less than 1, then f(x)=bx is decreasing for its entire domain.
In the graph, it is seen that f(x1)≥f(x2) if and only if x1≤x2. Considering the definition of f(x), the statement is proven.
Back home from the beach, Emily realized that she managed to solve an exponential equation to calculate the expiration of the chapati she and her friends cooked. However, she also realized that she has not practiced solving exponential inequalities. To help her practice, she went online to find some worksheets and found an interesting inequality.
Help Emily solve the inequality!Use the Exponential Property of Inequality.
a1=a-1
(am)n=am⋅n
aam=am−1
Write as a power
am⋅an=am+n
LHS+x<RHS+x
LHS−2<RHS−2
LHS/2<RHS/2
Put minus sign in front of fraction
Rearrange inequality
x | y=ln(x−2) | y=2logx | ||
---|---|---|---|---|
ln(x−2) | y | 2logx | y | |
1 | - | - | 2log1 | 0 |
2 | - | - | 2log2 | ≈0.6 |
3 | ln(3−2) | 0 | 2log3 | ≈0.95 |
4 | ln(4−2) | ≈0.69 | 2log4 | ≈1.2 |
5 | ln(5−2) | ≈1.1 | 2log5 | ≈1.4 |
12 | ln(12−2) | ≈2.3 | 2log12 | ≈2.16 |
Now both functions will be graphed on the same coordinate plane.
The number of solutions to the equation is the number of points of intersection of the graphs. These graphs have one point of intersection, so there is only one solution.
The solutions to the equation are the x-coordinates of any points of intersection of the graphs.
The x-coordinate of the point of intersection is about 8. Therefore, the solution to the equation is x≈8. This can be verified by substituting 8 for x in the given equation and checking whether a true statement is obtained.x≈8
Substitute -0.5 for pH and solve the logarithmic equation by graphing.
x | -logx | y |
---|---|---|
0.1 | -log0.1 | 1 |
0.5 | -log0.5 | ≈0.3 |
1 | -log1 | 0 |
2 | -log2 | ≈-0.3 |
3 | -log3 | ≈-0.48 |
Now, graph the functions on the same coordinate plane.
The graphs intersect at one point. The x-coordinate of the point of intersection is the hydrogen ion concentration of the solution.
The exact value of the x-coordinate cannot be identified. However, the answer must be rounded to the nearest mole per liter, so the hydrogen concentration of the solution with a pH of -0.5 is 3, rounded to the nearest mole per liter. The last step is to check the solution by substituting it back into the equation. Since a true statement was obtained, x≈3 is a solution to the equation.Let b be a positive real number different than 1. Two logarithms with the same base b are equal if and only if their arguments are equal.
logbx=logby⇔x=y
Since logarithms are defined for positive numbers, x and y must be positive.
The biconditional statement will be proved in two parts.
log2(m)+log2(n)=log2(mn)
logca=logbclogba
Calculate logarithm
LHS⋅2=RHS⋅2
m⋅log2(a)=log2(am)
(ab)m=ambm
LHS−x=RHS−x
Factor out x
Use the Zero Product Property
(II): LHS+1=RHS+1
(II): LHS/100=RHS/100