Problem 6 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, correct to four decimal places. $$ 3^{2 x-1}=5 $$

Step-by-Step Solution

Answer

The solution to the equation is approximately \(x = 1.2325\).

Step 1: Isolate the Exponential Expression

We begin with the equation \(3^{2x-1}=5\). The exponential expression, \(3^{2x-1}\), is already isolated on one side of the equation.

Step 2: Apply Logarithms to Both Sides

To solve for \(x\), apply the natural logarithm to both sides of the equation to get \(\ln(3^{2x-1})=\ln(5)\). This will allow us to bring down the exponent according to the logarithmic identity \(\log_b(a^n) = n \cdot \log_b(a)\).

Step 3: Use Logarithmic Identity

Apply the identity to simplify the left side: \((2x-1)\ln(3) = \ln(5)\). The equation now involves normal arithmetic terms.

Step 4: Solve for \(x\)

To isolate \(x\), first divide both sides by \(\ln(3)\), resulting in \(2x-1 = \frac{\ln(5)}{\ln(3)}\). Then, solve for \(x\) by adding 1 to both sides and then dividing by 2: \[x = \frac{\frac{\ln(5)}{\ln(3)} + 1}{2}\].

Step 5: Calculate the Numerical Solution

Calculate using a calculator: \(\ln(5) \approx 1.6094\) and \(\ln(3) \approx 1.0986\). First, compute \(\frac{1.6094}{1.0986} \approx 1.4650\). Next, add 1 to get \(1.4650 + 1 = 2.4650\) and then divide by 2 to find \(x \approx 1.2325\).

Key Concepts

Natural Logarithm

Natural logarithms are used to simplify exponential equations like the one given in the problem: \(3^{2x-1} = 5\). When dealing with exponential equations, natural logarithms (also known as base \(e\) logarithms, notated as \(\ln\)) are particularly useful. Converting from an exponential equation to a linear one becomes possible by applying the natural logarithm to both sides of the equation. This is because the natural logarithm and the exponential function are inverse operations.

• Applying the natural logarithm to an exponential expression allows us to move the exponent in front of the logarithm.
• This property will enable us to simplify and solve the equation more easily.

In our example, using the natural logarithm turns the equation into \(\ln(3^{2x-1}) = \ln(5)\), which is key to stripping away the complexity of our exponential term.

Logarithmic Identity

A logarithmic identity is a powerful tool when working with logarithms, especially when you have any expression in the form \(a^n\). The primary identity used in solving exponential equations is \(\log_b(a^n) = n \cdot \log_b(a)\). But don't be confused by the notation; here's how it works:

• The exponent \(n\) moves to the front of the logarithm, so you end up with a product rather than an exponentiated base.
• This transformation simplifies the equation from an exponential one to an arithmetic one, making it much easier to isolate variables such as \(x\).

Applying this identity, we turned \(\ln(3^{2x-1})\) into \((2x-1)\ln(3)\), allowing us to solve for \(x\) using basic algebraic manipulations.

Arithmetic Operations

Arithmetic operations are at the heart of solving the transformed logarithmic equations through basic algebra. Once the equation \((2x-1)\ln(3) = \ln(5)\) is established, we carry out several steps using basic arithmetic to find the value of \(x\):

• First, we divide both sides of the equation by \(\ln(3)\) to isolate the term \(2x-1\).
• Next, we simplify this to \(2x-1 = \frac{\ln(5)}{\ln(3)}\), making the equation ready for the final steps in solving for \(x\).
• By adding 1 to both sides and dividing by 2, we completely isolate \(x\).

These simple arithmetic operations help us compute the final value which, once calculated with a calculator, yields \(x \approx 1.2325\), correct to four decimal places, in our example. Mastering these basic steps enables you to tackle a wide range of exponential and logarithmic equations.