Problem 27 - Algebra 2

Question

Solve each equation or inequality. Round to four decimal places. $$ 2.1^{t-5}=9.32 $$

Step-by-Step Solution

Answer

\(t \approx 12.6236\)

Step 1: Isolate the Exponential Expression

The given equation is \(2.1^{t-5} = 9.32\). There is no need for isolation since the exponential expression \(2.1^{t-5}\) is already isolated on one side of the equation.

Step 2: Apply the Logarithm to Both Sides

To solve for \(t\), take the logarithm of both sides of the equation. This results in \(\log(2.1^{t-5}) = \log(9.32)\).

Step 3: Use Logarithmic Properties

Use the power rule for logarithms, which states that \(\log(a^b) = b \cdot \log(a)\). Apply this to \(\log(2.1^{t-5})\) to get \((t-5) \cdot \log(2.1) = \log(9.32)\).

Step 4: Solve for \(t - 5\)

Rearrange the equation to solve for \(t-5\). Divide both sides by \(\log(2.1)\) to get \(t-5 = \frac{\log(9.32)}{\log(2.1)}\).

Step 5: Calculate the Value

Calculate \(\frac{\log(9.32)}{\log(2.1)}\) using a calculator. This value is approximately \(7.6236\) when rounded to four decimal places.

Step 6: Solve for \(t\)

Add 5 to both sides of the equation from Step 5 to solve for \(t\). So, \(t = 7.6236 + 5\), which gives \(t \approx 12.6236\).

Key Concepts

Logarithmic Properties

Understanding logarithmic properties is crucial when solving exponential equations. A logarithm is the inverse operation of exponentiation, much like subtraction is the inverse of addition. One of the most common logarithmic properties is the power rule, which helps in breaking down complex expressions into simpler terms.

• The basic form is: if you have a number expressed as a power, like \( a^b \), you can express it logarithmically as \( b \cdot \log(a) \).
• Logarithms allow us to "bring down" exponents, which makes it easier to solve for variables tucked inside an exponent.

When you see an equation like \( 2.1^{t-5} = 9.32 \), you can apply logarithms to both sides to manage the base \( 2.1 \) and the exponent \( t-5 \). By understanding and utilizing these properties, solving equations becomes more straightforward.

Solving Equations

Solving equations involving exponential terms might seem tricky at first, but by taking systematic steps, it becomes manageable. In our exercise, the goal is to find the value of \( t \) in the equation \( 2.1^{t-5} = 9.32 \).

• First, ensure that the exponential expression is on one side to isolate it, though in our exercise, it is already isolated.
• Second, apply logarithms to both sides of the equation (\( \log(2.1^{t-5}) = \log(9.32) \)), which helps bring the exponent down using the power rule.
• Finally, one can solve for the variable by rearranging the equation and isolating the variable expression \( t - 5 \).

These steps are central to solving such equations and get us closer to finding the target variable. Make sure to perform operations carefully to avoid errors, especially when dealing with decimals or complex numbers.

Rounding Decimals

Rounding decimals correctly is an important skill in mathematics, especially when precision is necessary, as shown in our exercise. The rule of thumb for rounding is to look at the number directly to the right of the decimal place you are targeting.

• If that number is 5 or greater, round up the target number. If it's less than 5, you leave the target number as is.
• In our exercise, after calculating \( \frac{\log(9.32)}{\log(2.1)} \), you get a value of approximately \( 7.6236 \) when rounded to four decimal places, and when 5 is added, it becomes \( 12.6236 \).

Rounding helps ensure that calculations remain consistent and understandable without unnecessary complexity. Always double-check the rounded result to prevent any oversight, especially in academic work or assignments.

Power Rule of Logarithms

The power rule of logarithms is a powerful tool that simplifies the process of solving exponential equations. This rule states that \( \log(a^b) = b \cdot \log(a) \), allowing us to work more easily with equations involving powers.

• Using this property, you can take an exponent out of the logarithm, which reduces the problem to basic arithmetic operations.
• In our problem, applying the power rule allows us to transform \( \log(2.1^{t-5}) \) into \( (t-5) \cdot \log(2.1) \).
• With the exponent simplified, solving for the unknown variable becomes a linear process.

The power rule is essential when a variable is part of an exponent because it transforms complex expressions into more manageable ones. By mastering this rule, you can solve a wide range of mathematical problems more efficiently.