Problem 16 - Precalculus

Question

Solve the exponential equations. Make sure to isolate the base to a power first. Round our answers to three decimal places. $$15=7^{3-2 x}$$

Step-by-Step Solution

Answer

The solution is \( x \approx 0.804 \).

Step 1: Isolate the Exponential Expression

The given equation is \( 15 = 7^{3-2x} \). The exponential expression \( 7^{3-2x} \) is already isolated on one side of the equation. So this step is complete.

Step 2: Apply the Logarithm to Both Sides

To solve the equation \( 15 = 7^{3-2x} \), apply the natural logarithm to both sides to make use of the property that allows you to bring down the exponent: \[\ln(15) = \ln(7^{3-2x}).\] This simplifies to \[\ln(15) = (3-2x)\ln(7).\]

Step 3: Solve for the Exponent

From the equation \( \ln(15) = (3-2x)\ln(7) \), solve for \( x \). Start by dividing both sides by \( \ln(7) \): \[\frac{\ln(15)}{\ln(7)} = 3 - 2x.\] Rearrange this to solve for \( x \): \[2x = 3 - \frac{\ln(15)}{\ln(7)}.\] \[x = \frac{3 - \frac{\ln(15)}{\ln(7)}}{2}.\]

Step 4: Calculate the Value of x

Now calculate \( x \) using a calculator. Compute \( \ln(15) \approx 2.708 \) and \( \ln(7) \approx 1.946 \). \[\frac{\ln(15)}{\ln(7)} \approx \frac{2.708}{1.946} \approx 1.392.\] So the expression becomes: \[x = \frac{3 - 1.392}{2} = \frac{1.608}{2} = 0.804.\] Therefore, \( x \approx 0.804 \).

Key Concepts

Natural Logarithm

When faced with exponential equations like \(15 = 7^{3-2x}\), it's necessary to solve for the exponent. This is where the natural logarithm comes in handy. The natural logarithm, denoted as \(\ln\), is simply the logarithm to the base \(e\), where \(e\) is an irrational number approximately equal to 2.718. Using \(\ln\) helps us manage exponents in a systematic way.

The natural logarithm has a couple of key properties that are very useful. The most important one for solving our equation is \(\ln(a^b) = b \ln(a)\). This property allows you to "bring down" the exponent into a more manageable form. For our exercise, by applying \(\ln\) to both sides, we transformed the original equation into \(\ln(15) = (3-2x) \ln(7)\), making it easier to solve for \(x\).

This step is crucial as it converts the complex power operation into a simple linear equation.

Isolating the Base

Before you can solve an exponential equation, it's important to ensure the base of the expression is properly isolated. In our case, the equation \(15 = 7^{3-2x}\) already has the exponential term isolated, which means the \(7^{3-2x}\) part is alone on one side of the equation.

Isolating the base typically involves rearranging the equation such that the term containing the exponent is by itself on one side of the equals sign. This means moving any coefficients or other terms to the opposite side. Sometimes, it involves dividing or multiplying by constants to ensure the exponent term is isolated.

By isolating the base correctly, you simplify the equation and make it easier to apply logarithms and solve for the unknown variable. It sets the stage for effectively applying natural logarithms and manipulating the equation further.

Solving for Exponents

Solving for the exponent in an exponential equation like \(15 = 7^{3-2x}\) involves breaking it down into simpler, more manageable parts. After isolating the base and applying the natural logarithm, the equation becomes \(\ln(15) = (3-2x) \ln(7)\).

Here's where we solve for \(x\), the unknown in the exponent. First, divide both sides of the equation by \(\ln(7)\), which helps to isolate the term containing \(x\). This operation gives us \(\frac{\ln(15)}{\ln(7)} = 3 - 2x\).

Next, rearrange to solve for \(x\):

• Subtract the other terms from \(3\). This yields \(2x = 3 - \frac{\ln(15)}{\ln(7)}\).
• Divide everything by \(2\) to get \(x\). This simplifies to \[x = \frac{3 - \frac{\ln(15)}{\ln(7)}}{2}\].

Finally, you calculate the exact values using a calculator, resulting in \(x \approx 0.804\). By solving for the exponent in such a structured manner, you can handle a wide variety of exponential equations with ease.