Precalculus: Exponential and Logarithmic Functions
Solving, graphing, and modeling with exponentials and logarithms
Precalculus: Exponential and Logarithmic Functions
Solving, graphing, and modeling with exponentials and logarithms
Math - Grade 9-12
- 1
Evaluate f(x) = 3(2)^x for x = -2, x = 0, and x = 3.
The values are f(-2) = 3/4, f(0) = 3, and f(3) = 24 because 2^-2 = 1/4, 2^0 = 1, and 2^3 = 8. - 2
Rewrite each equation in the other form: log_5(125) = 3 and 10^-2 = 0.01.
A logarithm answers the question, what exponent is needed on the base.
The equation log_5(125) = 3 can be written as 5^3 = 125. The equation 10^-2 = 0.01 can be written as log_10(0.01) = -2. - 3
Solve for x: 2^(x + 1) = 16.
Since 16 = 2^4, the equation becomes 2^(x + 1) = 2^4. Therefore, x + 1 = 4 and x = 3. - 4
Solve for t: 7e^(0.4t) = 35.
After isolating the exponential expression, use the natural logarithm.
Divide both sides by 7 to get e^(0.4t) = 5. Taking the natural logarithm gives 0.4t = ln(5), so t = ln(5)/0.4, which is approximately 4.024. - 5
For the function y = log_2(x - 3) + 1, identify the domain, range, vertical asymptote, and the transformation from y = log_2(x).
Set the logarithm input x - 3 greater than zero.
The domain is x > 3, the range is all real numbers, and the vertical asymptote is x = 3. The graph of y = log_2(x) is shifted right 3 units and up 1 unit. - 6
Expand the expression using logarithm properties: ln(5x^3/sqrt(y)), where x and y are positive.
Write sqrt(y) as y^(1/2) before applying the power rule.
The expanded expression is ln(5) + 3ln(x) - 1/2ln(y). This uses the quotient rule, product rule, and power rule for logarithms. - 7
Condense the expression into a single logarithm: 2log(x) - (1/3)log(y) + log(4), where x and y are positive.
The condensed expression is log(4x^2/y^(1/3)). The coefficients become exponents, addition becomes multiplication, and subtraction becomes division. - 8
Solve for x and check for extraneous solutions: log_3(x - 2) + log_3(x + 2) = 2.
The logarithm inputs must both be positive before and after solving.
Combine the logarithms to get log_3((x - 2)(x + 2)) = 2, so x^2 - 4 = 9. Then x^2 = 13, so x = plus or minus sqrt(13). The domain requires x > 2, so the solution is x = sqrt(13). - 9
Find the inverse of f(x) = 4e^(x - 1) - 6.
Switch x and y after isolating the exponential part, or solve for x first and then rename the variable.
Start with y = 4e^(x - 1) - 6. Then y + 6 = 4e^(x - 1), so (y + 6)/4 = e^(x - 1). Taking the natural logarithm gives ln((y + 6)/4) = x - 1, so f^-1(x) = 1 + ln((x + 6)/4). - 10
An exponential function has the form y = ab^x and passes through the points (0, 5) and (3, 40). Find the values of a and b.
Using (0, 5), we get 5 = ab^0, so a = 5. Using (3, 40), we get 40 = 5b^3, so b^3 = 8 and b = 2. The function is y = 5(2)^x. - 11
A savings account starts with $1,200 and earns 4.5% annual interest compounded monthly. Write a model for the balance after t years, then find the balance after 6 years.
Use A = P(1 + r/n)^(nt), where n is the number of compounding periods per year.
The model is A(t) = 1200(1 + 0.045/12)^(12t). After 6 years, A(6) = 1200(1.00375)^72, which is approximately $1,570.14. - 12
A radioactive substance has a half-life of 9 days. If the initial amount is 80 grams, write a model for the amount A after t days and find the amount after 27 days.
The exponent t/9 counts how many half-lives have passed.
A half-life model is A(t) = 80(1/2)^(t/9). After 27 days, A(27) = 80(1/2)^3 = 10 grams. - 13
Order the functions ln(x), x, and 2^x from slowest growth to fastest growth as x becomes very large. Explain your reasoning.
The order from slowest to fastest growth is ln(x), x, and 2^x. Logarithmic functions grow more slowly than linear functions, and exponential functions eventually grow faster than any linear function. - 14
For the function y = 2^(-x) + 1, describe whether it represents growth or decay, find the horizontal asymptote, and state the y-intercept.
Rewrite 2^(-x) using a base between 0 and 1.
The function represents exponential decay because 2^(-x) is equivalent to (1/2)^x. The horizontal asymptote is y = 1, and the y-intercept is y = 2 because y = 2^0 + 1 = 2 when x = 0. - 15
Solve for x: 10^(2x - 1) = 7. Give an exact answer and a decimal approximation.
Taking log base 10 of both sides gives 2x - 1 = log(7). Therefore, x = (1 + log(7))/2, which is approximately 0.923.