Answer :
18. The formula for calculating the amount of money accumulated with continuous compounding is given by the formula:
A = P * e^(rt),
where A is the amount of money at the end of the investment period, P is the principal amount (initial investment), e is the base of the natural logarithm (approximately 2.71828), r is the interest rate, and t is the time period in years.
In this case, P = $10, r = 3% (or 0.03 as a decimal), and t = 15 years. Plugging in these values into the formula, we have:
A = 10 * e^(0.03 * 15).
Using a calculator or computer software, we can calculate this as:
A ≈ 10 * 2.22554.
Rounding to one decimal place, the amount of money at the end of 15 years is approximately $22.3.
19. To evaluate log4 32, we need to determine the exponent to which 4 must be raised to obtain 32. In other words, we want to solve the equation:
4^x = 32.
Taking the logarithm of both sides with base 4, we have:
log4 (4^x) = log4 32.
Using the property of logarithms that states log_b (b^x) = x, the equation simplifies to:
x = log4 32.
Using a calculator or computer software, we can evaluate this as:
x ≈ 2.5.
Therefore, log4 32 is approximately equal to 2.5.
20. The domain of the function g(x) = log3(3-3x) is determined by the argument of the logarithm. For the logarithm to be defined, the argument (3-3x) must be greater than zero. So, we need to solve the inequality:
3 - 3x > 0.
Simplifying this inequality, we have:
-3x > -3,
x < 1.
Therefore, the domain of the function g(x) is all real numbers less than 1.
21. To solve the equation 3x^2 + 2 = 27x + 4, we need to gather all the terms on one side and set the equation equal to zero:
3x^2 - 27x + 2 - 4 = 0,
3x^2 - 27x - 2 = 0.
Now, we can solve this quadratic equation by using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a),
where a, b, and c are the coefficients of the quadratic equation (ax^2 + bx + c = 0).
In this case, a = 3, b = -27, and c = -2. Substituting these values into the quadratic formula, we have:
x = (-(-27) ± √((-27)^2 - 4 * 3 * (-2))) / (2 * 3),
x = (27 ± √(729 + 24)) / 6,
x = (27 ± √753) / 6.
Therefore, the solutions to the equation are:
x ≈ 1.786 and x ≈ -5.786 (rounded to three decimal places).
22. To solve the equation log5 (2x - 1) - log5 (x - 2) = 1, we can use the properties of logarithms. The subtraction of logarithms is equivalent to the division of their arguments. Applying this property, we have:
log5 ((2x - 1)/(x
- 2)) = 1.
To eliminate the logarithm, we can rewrite the equation in exponential form:
5^1 = (2x - 1)/(x - 2).
Simplifying, we have:
5 = (2x - 1)/(x - 2).
Next, we can cross-multiply to eliminate the fraction:
5(x - 2) = 2x - 1.
Expanding and simplifying, we get:
5x - 10 = 2x - 1.
Bringing like terms to one side, we have:
5x - 2x = -1 + 10,
3x = 9.
Dividing by 3, we find:
x = 3.
Therefore, the solution to the equation is x = 3.
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