Problem 41 - Algebra and Trigonometry

Question

Solve each exponential equation . Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ 5^{2 x+3}=3^{x-1} $$

Step-by-Step Solution

Answer

The solution to the equation \(5^{2x+3}=3^{x-1}\) is approximately \(x = \frac{-3 \cdot ln(5) - ln(3)}{2 \cdot ln(5) - ln(3)}\). The decimal approximation should be calculated using a calculator.

Step 1: Transform the equation

Taking the natural logarithm \(ln\) on both sides of the equation \(5^{2x+3}=3^{x-1}\). We get:\[ln(5^{2x+3})=ln(3^{x-1})\]

Step 2: Use the properties of logarithms

Applying properties of logarithms, we can bring the exponents in front of the logs:\[(2x+3) \cdot ln(5)=(x-1) \cdot ln(3)\]

Step 3: Simplify and solve for x

By simplifying the above equation, we will isolate \(x\) on one side of the equation:\[2x \cdot ln(5) + 3 \cdot ln(5)=x \cdot ln(3) - ln(3)\]Moving all terms containing \(x\) to one side and constants to the other side, gives us:\[2x \cdot ln(5) - x \cdot ln(3) = -3 \cdot ln(5) - ln(3)\]Factorizing the left side to isolate \(x\):\[x \cdot (2 \cdot ln(5) - ln(3)) = -3 \cdot ln(5) - ln(3)\]Finally dividing both sides by \((2 \cdot ln(5) - ln(3))\) to obtain \(x\):\[x = \frac{-3 \cdot ln(5) - ln(3)}{2 \cdot ln(5) - ln(3)}\]

Step 4: Approximation of x

Use a calculator to obtain a decimal approximation for the solution, rounding to two decimal places. Check with a calculator if necessary.

Key Concepts

Natural Logarithms

Natural logarithms are a type of logarithms that use the number 'e' (approximately 2.71828) as their base. The natural logarithm of a number 'x' is usually written as \(ln(x)\) and represents the power to which 'e' must be raised to obtain the number 'x'. This concept is fundamental when dealing with exponential equations because the inverse operations of exponentiation with base 'e' are natural logarithms.

For example, \(ln(e^3) = 3\) because 'e' raised to the power of 3 returns back to the original number. When solving exponential equations, taking the natural logarithm of both sides is a common step because it allows us to 'bring down' the exponent, thereby linearizing the equation for easier solution. This is exactly what we utilize in our given exercise to solve for 'x'.

Properties of Logarithms

Logarithms have specific properties that make them extremely useful in algebra, particularly when solving exponential equations. Some of the key properties include:

• Product Rule: \(\ln(a \times b) = \ln(a) + \ln(b)\)
• Quotient Rule: \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
• Power Rule: \(\ln(a^b) = b \cdot ln(a)\)

The power rule is especially useful in our exercise, as it allows us to move the exponent in front of the logarithm. This transformation is critical to isolate the variable 'x' and solve the equation. The product and quotient rules can be used to further simplify expressions involving logarithms when handling more complex equations.

Exponential Equation with Different Bases

An exponential equation with different bases, like the one in our textbook exercise \(5^{2x+3} = 3^{x-1}\), presents a unique challenge. These types of equations cannot be easily solved by the same methods used for equations with like bases. Instead, we utilize logarithms to rewrite the exponents in a way that we can manage algebraically.

In our step-by-step example, applying natural logarithms to both sides allows us to leverage the aforementioned properties of logarithms to handle the different bases. By 'bringing down' the exponents and subsequently isolating 'x', we translate an exponential problem into a more familiar algebraic one. This is a crucial technique for solving exponential equations where direct comparison of bases is not possible.

Calculating Decimal Approximation

Once an algebraic solution for an equation is found, calculating a decimal approximation can provide a more concrete understanding of the solution's practical value. A decimal approximation is an estimate of an algebraic solution's value in decimal form, accurate to a certain number of decimal places. In the context of the given exercise, after solving for 'x' algebraically, we are left with a complex expression involving natural logarithms.

A calculator can then be used to evaluate this expression. Usually, we round the result to a specified number of decimal places for simplicity, which in our case is two decimal places. The decimal approximation not only gives us a usable number for practical applications but also serves as a means of verifying the correctness of our algebraic solution. It is important to check your work with a calculator to ensure the resulting approximation matches the expected range of the solution.