Problem 529 - Precalculus

Question

Find the exact solution for \(5^{2 x-3}=7^{x+1} .\) If there is no solution, write no solution.

Step-by-Step Solution

Answer

The exact solution for \(x\) is \( x = \frac{\ln(7) + 3 \cdot \ln(5)}{2 \cdot \ln(5) - \ln(7)} \).

Step 1: Apply logarithms to both sides

To eliminate the exponents, take the natural logarithm (ln) of both sides of the equation. \[ \ln(5^{2x-3}) = \ln(7^{x+1}) \]

Step 2: Use the power rule of logarithms

Apply the power rule, which states that \( \ln(a^b) = b \cdot \ln(a) \). This lets us rewrite the equation as follows:\[ (2x - 3) \cdot \ln(5) = (x + 1) \cdot \ln(7) \]

Step 3: Expand and simplify

Distribute the logarithm constants:\[ 2x \cdot \ln(5) - 3 \cdot \ln(5) = x \cdot \ln(7) + 1 \cdot \ln(7) \]

Step 4: Rearrange terms to isolate variable terms

Move all terms involving \(x\) to one side and constant terms to the other:\[ 2x \cdot \ln(5) - x \cdot \ln(7) = \ln(7) + 3 \cdot \ln(5) \]

Step 5: Factor out \(x\)

Factor \(x\) from the terms on the left:\[ x(2 \cdot \ln(5) - \ln(7)) = \ln(7) + 3 \cdot \ln(5) \]

Step 6: Solve for \(x\)

Divide both sides by the coefficient of \(x\):\[ x = \frac{\ln(7) + 3 \cdot \ln(5)}{2 \cdot \ln(5) - \ln(7)} \]

Key Concepts

Logarithms

Logarithms are a mathematical tool that helps simplify equations involving exponents. If you have an equation that deals with exponential terms, you can use logarithms to linearize the equation, which makes finding solutions easier. The key property of logarithms is that they can transform a multiplicative relationship into an additive one. This is because the logarithm of a product is the sum of the logarithms:

• For example, \( \log(a \cdot b) = \log(a) + \log(b) \)
• Also, the logarithm of a power is the exponent times the logarithm of the base: \( \log(a^b) = b \cdot \log(a) \)

Additionally, utilizing the properties of logarithms such as the base, power, and division rules can greatly ease the process of solving exponential and logarithmic equations.

Power Rule of Logarithms

The Power Rule of Logarithms is an invaluable tool when dealing with equations where exponents are present in the terms. This rule states that the logarithm of a number raised to an exponent is equal to the exponent multiplied by the logarithm of the base number, expressed as:

• \( \log_b(a^c) = c \cdot \log_b(a) \)

This allows us to take the exponent "down" in front of the log, converting a multiplication problem into a simpler pair of terms. This transformation simplifies complex equations, making them easier to manage and solve.
When using this rule, be sure the base of the logarithm and the base of the exponent are compatible or adjusted correctly, so calculations remain consistent. The Power Rule is especially powerful in algebra when dealing with complicated expressions or solving exponential equations.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is logarithm with a base of \( e \), where \( e \) is an irrational mathematical constant approximately equal to 2.71828. It appears in many areas of calculus and is often used when dealing with problems involving growth and rates of change.
In mathematics, natural logarithms are preferred because they have special properties that make calculations easier, especially in calculus. For example:

• The derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \)
• The integral of \( \ln(x) \) can be found using the technique of integration by parts

Using natural logarithms in exponential equations, like the one in the original exercise, allows to easily apply the power rule to simplify and rearrange the equation. This step makes it possible to extract the variable from the exponent, easing the solving process.

Solving Equations

Solving equations, especially when exponents are involved, can sometimes seem complex. The key is to systematically approach and simplify the problem:

• Begin by examining the equation and identifying what kind it is, such as exponential, logarithmic, or polynomial
• Next, look for suitable methods or transformations that help simplify the equation, like using logarithms to handle exponents
• In our exercise, applying the natural logarithm allowed us to bring the exponents down, so they became ordinary linear terms
• After simplifying, group similar terms together, isolating variables when possible
• Finally, solve for the variable step by step, checking the solution's validity in the original context of the problem

By following these steps, the task of finding solutions becomes manageable and is a useful strategy for approaching many different types of equations. Always double-check solutions to ensure they fit the parameters of the original equation.