High School

Section 1.5

18. If $10 is invested for 15 years at 3% interest compounded continuously, find the amount of money at the end of 15 years. Answer correct to one decimal place. 19. Evaluate log4 32 20. Find the domain of the function g(x) = log3(3-3x)

21. Solve the equation 3x2+2 = 27x+4

22. Solve the equation log5 (2x-1)-log5 (x-2)= 1

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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