Problem 74 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ 29=5^{x-6} $$

Step-by-Step Solution

Answer

Exact: \(x = \frac{\ln(29)}{\ln(5)} + 6\); Approximate: \(x \approx 8.0911\).

Step 1: Set the Stage by Applying the Logarithm

To solve the equation \(29 = 5^{x-6}\), first take the natural logarithm of both sides to facilitate the solving for \(x\). The equation becomes \(\ln(29) = \ln(5^{x-6})\).

Step 2: Apply the Power Rule for Logarithms

Use the power rule for logarithms which states that \(\ln(a^b) = b \cdot \ln(a)\). Apply this rule to the right side of the equation: \(\ln(29) = (x - 6) \ln(5)\).

Step 3: Solve for \(x\)

To isolate \(x\), divide both sides of the equation by \(\ln(5)\): \[ x - 6 = \frac{\ln(29)}{\ln(5)} \]Then add 6 to both sides to solve for \(x\): \[ x = \frac{\ln(29)}{\ln(5)} + 6 \].

Step 4: Compute the Exact Solution

Use a calculator to compute \(\frac{\ln(29)}{\ln(5)}\) to find the exact value. This results in approximately 2.116 or exactly \(\frac{\ln(29)}{\ln(5)}\). Add 6 to get the final exact value of \(x\):\[ x = \frac{\ln(29)}{\ln(5)} + 6 \].

Step 5: Calculate the Approximate Decimal Value

Calculate the decimal approximation using a calculator. Substitute the decimals for \(\ln(29)\) and \(\ln(5)\) to find:\[ x \approx \frac{3.3673}{1.6094} + 6 \approx 2.0911 + 6 \approx 8.0911 \].Thus, the approximate value of \(x\) is 8.0911.

Key Concepts

Solving Equations

When faced with an exponential equation like \(29 = 5^{x-6}\), the main goal is to solve for \(x\). The process involves a few systematic steps that make the equation easier to handle.

• First, we need to "de-exponentiate" the expression, so to speak. This means getting \(x\) out of the exponential form.
• This is where logarithms, which are the inverses of exponentials, come into play.
• By taking the logarithm of both sides of the equation, we can rewrite the expression in a way that uncovers \(x\).

After applying the logarithm, we use algebraic manipulation to isolate and solve for \(x\). This often involves applying the power rule for logarithms which simplifies the expression further. Through this process, we systematically solve for \(x\) by treating it much like solving a standard linear equation, but with logarithmic terms.

Logarithms

Logarithms are a key tool used to transition between exponential and linear forms of equations. Here are some vital concepts that revolve around logarithms:

• Logarithms can be thought of as the "undoing" of an exponential function. While exponentials involve raising a base to a power, logarithms ask the question "to what power must this base be raised to produce a particular number?"
• The natural logarithm, \(\ln\), is the logarithm with base \(e\), where \(e\approx2.718\).
• In our equation, \(\ln(29)\) and \(\ln(5)\) are crucial because they help translate the exponential equation into a more manageable form.
• An essential rule to remember is the power rule: \( \ln(a^b) = b \cdot \ln(a) \).

In our exercise, after logarithms are applied to both sides, we used the power rule to derive a linear equation, \(\ln(29) = (x - 6)\ln(5)\), which can be easily manipulated to find \(x\). This demonstrates how logarithms are instrumental in dealing with exponential equations.

Exact and Approximate Solutions

There are often two types of solutions we aim for with equations: exact and approximate. Each serves a different purpose.

• The exact solution is usually left in a form that involves fractions, square roots, or, as in our case, logarithms. This represents the answer in the most precise mathematical terms possible.
• For our equation, the exact solution came out as \[ x = \frac{\ln(29)}{\ln(5)} + 6 \].
• Approximate solutions, meanwhile, provide numerical estimates. They're particularly useful when we need a quick value for practical applications or where only decimal results are appropriate.

Upon computing the decimals for \(\ln(29)\) and \(\ln(5)\), we achieved an approximate solution: \(x \approx 8.0911\). This approximation is crucial in real-life scenarios where we need an answer in simple figures. It is always important to know how to arrive at both forms, as each is useful in different contexts.