Questions: Solve the equation. 4 + log(8x) = 25 - 2 log(x) x = 5000000 Solve the equation. ln(x + 4) + ln(x) = ln(x + 40) x = 5 The sales decay for a product is given by S = 80,000 e^(-0.5x), where S is the monthly sales and x is the number of months that have passed since the end of a promotional campaign. (a) What will be the sales 2 months after the end of the campaign? (Round your answer to two decimal places.) (b) How many months after the end of the campaign will sales drop below 1,000, if no new campaign is initiated? (Round up to the nearest whole number.) The demand function for a certain commodity is given by p = 100 e^(-q / 2). (p is the price per unit and q is the number of units.) (a) At what price per unit will the quantity demanded equal 8 units? (Round your answer to the nearest cent.) (b) If the price is 2.83 per unit, how many units will be demanded, to the nearest unit? The total cost function for a product is C(x) = 600 ln(x + 10) + 1900 where x is the number of units produced. (a) Find the total cost of producing 100 units. (Round your answer to the nearest cent.) (b) Producing how many units will give total costs of 6500? (Round your answer to the nearest whole number.)

1. Combine the logarithmic terms on one side of the equation.
2. Use properties of logarithms to simplify the equation.
3. Solve for \(x\).

First, we need to isolate the logarithmic terms. Let's move all the logarithmic terms to one side of the equation:

\[ 4 + \log(8x) + 2\log(x) = 25 \]

Next, we use the properties of logarithms to combine the terms:

\[ 4 + \log(8x) + \log(x^2) = 25 \]

\[ 4 + \log(8x \cdot x^2) = 25 \]

\[ 4 + \log(8x^3) = 25 \]

Subtract 4 from both sides:

\[ \log(8x^3) = 21 \]

Rewrite the logarithmic equation in exponential form:

\[ 8x^3 = 10^{21} \]

Solve for \(x\):

\[ x^3 = \frac{10^{21}}{8} \]

\[ x^3 = 1.25 \times 10^{20} \]

\[ x = \sqrt[3]{1.25 \times 10^{20}} \]

\[ x \approx 5 \times 10^6 \]

\[ \boxed{x \approx 5000000} \]

First, use the properties of logarithms to combine the terms on the left side:

\[ \ln((x+4)x) = \ln(x+40) \]

Simplify the argument of the logarithm:

\[ \ln(x^2 + 4x) = \ln(x+40) \]

Since the natural logarithm function is one-to-one, we can equate the arguments:

\[ x^2 + 4x = x + 40 \]

Rearrange the equation to form a quadratic equation:

\[ x^2 + 3x - 40 = 0 \]

Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-40)}}{2(1)} \]

\[ x = \frac{-3 \pm \sqrt{9 + 160}}{2} \]

\[ x = \frac{-3 \pm \sqrt{169}}{2} \]

\[ x = \frac{-3 \pm 13}{2} \]

This gives us two solutions:

\[ x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{-16}{2} = -8 \]

Since \(x\) must be positive (as it is inside a logarithm), we discard \(x = -8\):

\[ \boxed{x = 5} \]

Given the sales decay function \(S = 80,000 e^{-0.5x}\):

Substitute \(x = 2\) into the function:

\[ S = 80,000 e^{-0.5 \times 2} \]

\[ S = 80,000 e^{-1} \]

Using the value of \(e^{-1} \approx 0.3679\):

\[ S \approx 80,000 \times 0.3679 \]

\[ S \approx 29,432 \]

\[ \boxed{S \approx 29,432} \]

Set \(S = 1,000\) and solve for \(x\):

\[ 1,000 = 80,000 e^{-0.5x} \]

Divide both sides by 80,000:

\[ \frac{1,000}{80,000} = e^{-0.5x} \]

\[ 0.0125 = e^{-0.5x} \]

Take the natural logarithm of both sides:

\[ \ln(0.0125) = -0.5x \]

\[ -4.3820 = -0.5x \]

Solve for \(x\):

\[ x = \frac{4.3820}{0.5} \]

\[ x \approx 8.764 \]

Rounding up to the nearest whole number:

\[ \boxed{x = 9 \text{ months}} \]