Problem 22 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, correct to four decimal places. $$ 7^{x / 2}=5^{1-x} $$

Step-by-Step Solution

Answer

\( x \approx -0.7526 \)

Step 1: Take the Natural Logarithm of Both Sides

To solve the equation \( 7^{x/2} = 5^{1-x} \), we start by taking the natural logarithm of both sides:\[ \ln(7^{x/2}) = \ln(5^{1-x}) \] Using the property \( \ln(a^b) = b \ln(a) \), we can simplify both sides.

Step 2: Apply Logarithm Power Rule

Using the power rule of logarithms, rewrite the logs based on the powers:\[ \frac{x}{2} \ln(7) = (1-x) \ln(5) \]This simplifies the equation and allows us to solve for \( x \).

Step 3: Distribute and Rearrange

Distribute the logarithm terms to open them up:\[ \frac{x}{2} \ln(7) = \ln(5) - x \ln(5) \]Now, move all terms involving \( x \) to one side of the equation:\[ \frac{x}{2} \ln(7) + x \ln(5) = \ln(5) \]

Step 4: Factor Out \( x \)

Factor out \( x \) on the left side:\[ x \left( \frac{\ln(7)}{2} + \ln(5) \right) = \ln(5) \]

Step 5: Solve for \( x \)

Divide both sides by the expression in parenthesis to solve for \( x \):\[ x = \frac{\ln(5)}{\frac{\ln(7)}{2} + \ln(5)} \]Calculate the value using a calculator and round to four decimal places.

Step 6: Calculate the Final Result

Using a calculator, compute:1. Calculate \( \ln(7) \) and \( \ln(5) \).2. Compute \( \frac{\ln(7)}{2} \).3. Sum \( \frac{\ln(7)}{2} + \ln(5) \).4. Divide \( \ln(5) \) by the result from step 3.The result, rounded to four decimal places, is \( x \approx -0.7526 \).

Key Concepts

Natural Logarithm

When you encounter equations with exponential terms, the natural logarithm, or "ln," becomes a useful tool. The natural logarithm is simply the logarithm to the base of the mathematical constant \( e \), which is approximately equal to 2.71828. By taking the natural logarithm of both sides of an equation, you can transform these exponential expressions into a format that's easier to work with.

In the exercise, starting with \( 7^{x/2} = 5^{1-x} \), we apply \( \ln \) to both sides. This action leverages the unique properties of logarithms that help simplify the equation into terms that allow further algebraic manipulation. The application of \( \ln \) helped transition from exponentials to a linear form of \( \ln(7^{x/2}) = \ln(5^{1-x}) \).

This transformation is crucial because it opens the door for using logarithmic properties to further simplify and eventually solve the equation.

Logarithm Power Rule

The logarithm power rule is vital when simplifying exponential equations. It states that for any positive numbers \( a \) and \( b \), the natural logarithm of \( a^b \) is equal to \( b \ln(a) \). This rule is represented in the formula:

• \( \ln(a^b) = b \ln(a) \)

The power rule allows us to move the exponent out in front of the logarithm, thereby converting complex exponential expressions into manageable linear terms.

With the exercise's equation \( 7^{x/2} = 5^{1-x} \), by applying the power rule, we reduce it to \( \frac{x}{2} \ln(7) = (1-x) \ln(5) \).

This step is crucial; it converts the original exponential terms into linear forms, making it much easier to isolate and solve for the variable \( x \). Understanding and executing the power rule correctly is essential when working with logarithms in exponential equations.

Solve for x

Solving for \( x \) in an equation means isolating \( x \) on one side to determine its value. In this process, we use the transformed linear equations obtained after applying logarithmic properties. After simplifying our equation using the logarithm power rule, we arrive at:

• \( \frac{x}{2} \ln(7) + x \ln(5) = \ln(5) \)

At this point, our goal is to consolidate all \( x \)-related terms together. This is done by factoring \( x \) out:

• \( x \left( \frac{\ln(7)}{2} + \ln(5) \right) = \ln(5) \)

This makes \( x \) the subject of the equation, as everything now hinges on computing that expression.

Finally, divide both sides by the factor to determine \( x \):

• \( x = \frac{\ln(5)}{\frac{\ln(7)}{2} + \ln(5)} \)

Performing these steps accurately ensures that the variable \( x \) can be found correctly and in this case, its value is approximately \(-0.7526\), rounded to four decimal places.

Algebraic Manipulation

Algebraic manipulation is at the heart of solving equations. It involves rearranging equations to isolate variables, using properties like distribution and factoring. In our context, once we identified an expression involving \( x \), the next step was to simplify and rearrange everything logically.

Starting from \( \frac{x}{2} \ln(7) = \ln(5) - x \ln(5) \), we needed to ensure all \( x \)-terms appeared on one side. By moving terms and factoring \( x \), we obtain:

• \( x \left( \frac{\ln(7)}{2} + \ln(5) \right) = \ln(5) \)

This strategic manipulation, through distribution and rearrangement, is crucial. It facilitates expressing \( x \) in a form directly ready for computation.

Finally, by executing the division to solve for \( x \), you engage in essential algebraic manipulation. This is a fundamental tool along with using calculators to evaluate expressions, achieving the target result.