Problem 19 - College Algebra

Question

For the following exercises, use logarithms to solve. $$ -6 e^{9 x+8}+2=-74 $$

Step-by-Step Solution

Answer

The solution for \(x\) is approximately \(-0.342\).

Step 1: Isolate the Exponential

First, we start by isolating the exponential term in the equation. The given equation is \[-6 e^{9x+8} + 2 = -74\] Subtract 2 from both sides to isolate the exponential term:\[-6 e^{9x+8} = -76\]

Step 2: Divide to Simplify

Divide both sides by -6 to further isolate the exponential expression:\[e^{9x+8} = \frac{76}{6}\] Simplify the fraction:\[e^{9x+8} = \frac{38}{3}\]

Step 3: Take the Natural Logarithm

Apply the natural logarithm to both sides of the equation to eliminate the exponential function:\[\ln(e^{9x+8}) = \ln\left(\frac{38}{3}\right)\] Using the property of logarithms, \(\ln(e^y) = y\), simplify the left side:\[9x + 8 = \ln\left(\frac{38}{3}\right)\]

Step 4: Solve for x

Now, we solve for \(x\) by isolating it on one side of the equation. Subtract 8 from both sides:\[9x = \ln\left(\frac{38}{3}\right) - 8\] Then, divide by 9:\[x = \frac{\ln\left(\frac{38}{3}\right) - 8}{9}\]

Step 5: Final Calculation Step

Use a calculator to evaluate the expression for \(x\):\(\ln\left(\frac{38}{3}\right)\) Calculate \(\ln\left(\frac{38}{3}\right)\) and then substitute back into the equation:\[x = \frac{\ln\left(\frac{38}{3}\right) - 8}{9} \approx -0.342\]

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical operations that help in solving exponentials by reversing the operation of exponentiation. They are essential in cases where you are handling equations involving exponential expressions. Logarithms can transform multiplicative processes into additive ones, simplifying complexities.

• Logarithmic functions are typically written as \( \log_b(x) \), where \(b\) is the base and \(x\) is the argument.
• They answer the question: To what power must the base \(b\) be raised to produce \(x\)?
• For example, \( \log_2(8) = 3 \) because \(2^3 = 8\).

In solving exponential equations, logarithms can be applied to both sides of the equation. This method permits the transition from an exponential equation to a linear one. Hence, once a logarithm of the same base as the exponent is taken, the power can be brought down in front as a multiplier, effectively making the equation easier to manage and solve.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a specific type of logarithm where the base is the mathematical constant \( e \) (approximately equal to 2.71828). It is a crucial concept when dealing with continuous growth or decay rates, making it prevalent in many scientific and financial calculations.

• The natural logarithm of a number is the power to which \( e \) must be raised to obtain that number.
• For example, \( \ln(e) = 1 \) since \( e^1 = e \).

In the exercise solution, the natural logarithm plays a significant role when dealing with the expression \( e^{9x+8} \). Applying the natural logarithm, \( \ln \), to both sides allows you to utilize the property \( \ln(e^y) = y \). This property simplifies the calculation by removing the exponential nature of the equation, hence making the framework linear and easier to solve.

Exponential Functions

Exponential functions are expressions in which a variable represents the exponent of a fixed base. These functions exhibit mathematical operations involving constant ratio growth or decay, and they appear frequently in diverse scientific fields such as biology, chemistry, economics, and physics.

• An exponential function is usually expressed in the form \( f(x) = a \cdot b^x \), where \( b \) is a positive real number and \( x \) is the variable.
• They are characterized by showing a rapid increase or decrease.

In the scenario given in the exercise, \( e^{9x+8} \) is the exponential component we are solving for. To make such an equation solvable, the first step involves isolating the exponential part onto one side of the equation, as seen in the steps provided. By doing so, it enables the subsequent application of logarithmic functions which transform the exponential problem into a simpler, linear one, flowing naturally into the solution.