Problem 12 - Contemporary Precalculus

Question

Solve the equation. First express your answer in terms of natural logarithms (for instance, \(x=(2+\ln 5) /(\ln 3)) .\) Then use a calculator to find an approximation for the answer. $$9^{x-1}=8^{x-3}$$

Step-by-Step Solution

Answer

- The equation was $$9^{x-1}=8^{x-3}$$. 2. What logarithm was applied to both sides of the equation? - The natural logarithm (ln) was applied to both sides of the equation. 3. What is the simplified equation for x before finding the numerical approximation? - The simplified equation for x before finding the numerical approximation is: $$x=\frac{\ln9-3\ln8}{\ln9-\ln8}$$. 4. What is the approximate numerical value of x? - The approximate numerical value of x is 4.942.

Step 1: Rewrite the equation using properties of exponents

The equation can be rewritten as: $$9^{x}\cdot9^{-1}=8^{x}\cdot8^{-3}$$

Step 2: Apply natural logarithm to both sides of the equation

Applying natural logarithm, we get: $$\ln(9^x\cdot9^{-1})=\ln(8^x\cdot8^{-3})$$

Step 3: Simplify and isolate the variable x

Utilize the properties of logarithms to separate the terms with x: $$x\ln9-\ln9=x\ln8-3\ln8$$ Now, isolate the variable x: $$x\ln9-x\ln8=\ln9-3\ln8$$ Factor the x out of the equation: $$x(\ln9-\ln8)=\ln9-3\ln8$$ Divide both sides by \((\ln9-\ln8)\) to get x: $$x=\frac{\ln9-3\ln8}{\ln9-\ln8}$$

Step 4: Compute the numerical approximation

Use a calculator to find the approximately numerical value of x: $$x\approx 4.942$$ So the solution to the equation is \(x\approx 4.942\).

Key Concepts

Natural Logarithms

Natural logarithms are a crucial tool when working with exponential equations. When we refer to the natural logarithm, we are using a logarithm with the base of the irrational constant, Euler's number, denoted as \(e\). This special number is roughly equal to 2.71828. Natural logarithms are denoted as \(\ln\), so \(\ln(x)\) represents the logarithm of \(x\) to the base \(e\).

Using natural logarithms is particularly helpful because they simplify calculations when dealing with exponential growth or decay in natural processes. They have unique properties that make them efficient for solving equations involving exponents. For instance, the natural logarithm allows us to "bring down" the exponent in an equation, making it easier to manage algebraically. This property is vital in solving equations like \(9^{x-1}=8^{x-3}\) as it helps isolate the variable \(x\).

Properties of Logarithms

The properties of logarithms are incredibly useful, especially when solving equations involving exponents. Understanding these properties enables us to manipulate and simplify expressions effectively. Here are some key properties:

• **Product Property**: \(\ln(ab) = \ln a + \ln b\) - This property is useful for breaking down multiplicative expressions.
• **Quotient Property**: \(\ln(\frac{a}{b}) = \ln a - \ln b\) - This helps in separating fractions into simpler terms.
• **Power Property**: \(\ln(a^b) = b\ln a\) - This is particularly powerful as it allows exponents to be transformed into coefficients, simplifying equations.

In our problem, we use the power property by applying the natural logarithm to both sides of the transformed equation. By doing this, we transfer the exponent \(x\) into a multiplicative factor, \(x\ln\), simplifying the process of isolating \(x\) and solving the equation.

Properties of Exponents

Exponents play a fundamental role in algebra and are represented as a number raised to the power of another number. They describe a repeated multiplication of a base. Here are essential properties that were applied:

• **Product of Powers**: \(a^m \cdot a^n = a^{m+n}\) - Used when multiplying like bases.
• **Quotient of Powers**: \(\frac{a^m}{a^n} = a^{m-n}\), when \(a eq 0\) - Used when dividing like bases.
• **Power of a Power**: \((a^m)^n = a^{m\cdot n}\) - When raising an exponential expression to another power.

In solving our original equation, \(9^{x-1} = 8^{x-3}\), the exponent properties allow rewriting the expression as \(9^x \cdot 9^{-1}\) and similarly for the base 8. This transformation makes it more convenient to apply logarithmic properties later, simplifying the steps towards finding \(x\).

Solving Algebraic Equations

When solving algebraic equations, our goal is to find the value of unknown variables that satisfy the equation. The problem \(9^{x-1}=8^{x-3}\) is an example of an exponential equation, where variables appear in the exponent.

• **Rewriting Equations**: As shown, transforming the equation to manageable terms using exponent properties makes solving easier.
• **Isolating the Variable**: Use operations like addition, subtraction, multiplication, or division to get your variable alone on one side of the equation.
• **Using Logarithms**: Especially for exponential equations, applying logarithms can demystify exponent variables, making them tangible coefficients.

In our exercise, we first rewritten the problem using exponent properties and then utilized natural logarithms to simplify and isolate \(x\). Finally, we solved for \(x\) algebraically, resulting in a clear, approximate numerical solution.