Problem 15 - Precalculus

Question

Solve the exponential equations. Make sure to isolate the base to a power first. Round our answers to three decimal places. $$27=2^{3 x-1}$$

Step-by-Step Solution

Answer

The solution is approximately \( x \approx 1.919 \).

Step 1: Write the Equation

The given exponential equation is \( 27 = 2^{3x-1} \).

Step 2: Take the Logarithm of Both Sides

To solve for \( x \), take the natural logarithm on both sides: \[ \ln(27) = \ln(2^{3x-1}) \]

Step 3: Use the Power Rule for Logarithms

Apply the power rule, which states \( \ln(a^b) = b\ln(a) \), to simplify the right side: \[ \ln(27) = (3x-1)\ln(2) \]

Step 4: Solve for x

First, divide both sides by \( \ln(2) \) to isolate \( 3x-1 \): \[ \frac{\ln(27)}{\ln(2)} = 3x - 1 \] Next, add 1 to both sides to get: \[ \frac{\ln(27)}{\ln(2)} + 1 = 3x \] Finally, divide by 3 to solve for \( x \): \[ x = \frac{1}{3} \left( \frac{\ln(27)}{\ln(2)} + 1 \right) \].

Step 5: Calculate the Numerical Value of x

Use a calculator to compute the values: \( \ln(27) \approx 3.295837 \) and \( \ln(2) \approx 0.693147 \). Now substitute these values into the equation for \( x \): \[ x = \frac{1}{3} \left( \frac{3.295837}{0.693147} + 1 \right) \approx \frac{1}{3} (4.7568 + 1) \approx \frac{1}{3} (5.7568) \approx 1.919 \].

Key Concepts

Logarithms

Logarithms are powerful tools to help simplify complex equations, especially when dealing with exponential expressions. In exponential equations like \(27 = 2^{3x-1}\), it can be difficult to directly solve for \(x\) because the variable \(x\) is in the exponent. To manage this, we take the logarithm of both sides of the equation. This is allowed because whatever you do to one side of the equation, you do to the other to keep it balanced.

• Taking the logarithm of both sides helps to bring the exponent, containing the variable, down to a level where regular algebraic techniques can be used.
• In this problem, we used the natural logarithm, denoted as \(\ln\), which is quite common in mathematics.
• The process is beneficial because logarithms transform multiplication into addition and exponents into multiplication, making the equation simpler to manipulate.

It is crucial to realize that utilizing logarithms is a standard method for solving equations where the unknown is in the exponent, making them indispensable for such problems.

Power Rule

The power rule for logarithms is a fundamental tool used when solving equations that have the variable in an exponent. The power rule states that \(\ln(a^b) = b \ln(a)\). This means you can "bring down" the exponent as a coefficient in front of the logarithm, which makes the equation much simpler.

• By applying the power rule in our exercise, \(\ln(2^{3x-1})\) becomes \((3x-1)\ln(2)\).
• This transforms the complicated exponential equation into a straightforward algebraic equation.

Remember, the power rule is pivotal because it allows us to separate the exponent from the base, facilitating easier manipulation of the equation to where traditional algebraic operations can be employed. Without this rule, solving such equations would be cumbersome and difficult.

Solving for x

Once we've transformed the equation using logarithms and the power rule, the next step is to solve for \(x\). With the variable \(x\) out of the exponent, we can follow traditional algebraic steps to isolate it.

• First, we divide both sides of the equation by \(\ln(2)\), in order to isolate the bracket \((3x-1)\) on one side.
• Then, we add 1 to both sides to move constants over and simplify the equation.
• Finally, dividing everything by 3 will fully isolate \(x\), giving us the solution for the variable.

This sequential process highlights the importance of algebraic manipulation after using logarithms to transform the problem. These steps, once executed correctly, lead us to the final answer for \(x\), which can be rounded appropriately as per the problem's requirements. Mastering these techniques is essential for efficiently solving exponential equations.