Problem 12 - Algebra 2 and Trigonometry

Question

In \(3-14,\) solve each equation for the variable. Express each answer to the nearest hundredth. $$ (5 \times 7)^{a}=0.585 $$

Step-by-Step Solution

Answer

The value of \(a\) is approximately -0.15.

Step 1: Simplify the Base

First, simplify the expression inside the parenthesis. Calculate \(5 \times 7\). This evaluates to 35, so the equation becomes \(35^a = 0.585\).

Step 2: Apply Logarithms

To solve the equation \(35^a = 0.585\), apply the natural logarithm to both sides to use the power rule of logarithms. This gives us \(\ln(35^a) = \ln(0.585)\). By the logarithmic identity, this is equivalent to \(a \cdot \ln(35) = \ln(0.585)\).

Step 3: Isolate the Variable

Solve for \(a\) by dividing both sides of the equation by \(\ln(35)\). This results in \(a = \frac{\ln(0.585)}{\ln(35)}\).

Step 4: Calculate the Value of \(a\)

Substitute the natural log values into the equation: \(a = \frac{\ln(0.585)}{\ln(35)}\). Using a calculator, find \(\ln(0.585)\) and \(\ln(35)\).

Step 5: Round to Nearest Hundredth

Calculate \(\ln(0.585) \approx -0.535\) and \(\ln(35) \approx 3.555\). Substitute back: \(a = \frac{-0.535}{3.555} \approx -0.150.\) Round this to the nearest hundredth to get \(a \approx -0.15\).

Key Concepts

Understanding the Natural Logarithm

The natural logarithm, often represented by \(\ln\), is a special kind of logarithm that uses the mathematical constant \(e\) as its base. The value of \(e\) is approximately 2.718. The natural logarithm is very useful in calculus and helps us to solve equations involving exponential terms. When you see \(\ln(x)\), it’s asking the question: "To what power must \(e\) be raised, to equal \(x\)?"
For example, if \(\ln(35)\) is calculated, we're looking for the power that \(e\) must be raised to, in order to get 35. Understanding \(\ln\) is crucial as it converts multiplicative problems into additive ones, making them easier to solve.

Solving Equations with Logarithms

When given an equation with exponential terms, such as \(35^a = 0.585\), it may not be straightforward to solve directly. This is where logarithms come in handy.
The power rule of logarithms states that the logarithm of a power can be rewritten as a product: \(\ln(b^c) = c \cdot \ln(b)\). Applying this rule helps isolate the variable.

• Take the natural logarithm of both sides: \(\ln(35^a) = \ln(0.585)\).
• This transforms to: \(a \cdot \ln(35) = \ln(0.585)\).

This simplifies the process and allows you to solve for \(a\) by dividing both sides by \(\ln(35)\).

The Role of Exponentiation

Exponentiation is the operation of raising one number, the base, to the power of another number, the exponent. In the equation \(35^a = 0.585\), 35 is the base and \(a\) is the exponent. This operation is central to many mathematical expressions, especially those involving growth, decay, or interest.
Understanding how exponents work helps in setting up the problems correctly and choosing the right method to find solutions. In this particular problem, the challenge is to "undo" the exponentiation affecting the base 35. Using logarithms makes this possible as it brings the exponent down as a multiplicative factor, simplifying the equation and making it solvable.

Rounding Decimals to the Nearest Hundredth

Rounding decimals is a critical step in presenting your final answer accurately and clearly. To round a number to the nearest hundredth, you look at the thousandth place (the third digit after the decimal).

• If this digit is 5 or greater, round up the last number in the hundredth place.
• If the digit is less than 5, leave the number in the hundredth place unchanged.

In our equation solution, after computing \(a \approx -0.150\), we round it to \(-0.15\) because the third decimal place digit "0" is less than "5". Rounding ensures numbers are manageable and consistent, especially important in scientific and financial calculations.