Problem 80 - A Graphical Approach to College Algebra

Question

$$\text { Solve each formula for the indicated variable.}$$ $$D=160+10 \log x, \text { for } x$$

Step-by-Step Solution

Answer

x = 10^{\frac{D - 160}{10}}

Step 1: Isolate the Logarithmic Term

Start by isolating the term that contains the logarithm. This means you need to move the constant term to the other side of the equation. You accomplish this by subtracting 160 from both sides of the equation: \[ D - 160 = 10 \log x \]

Step 2: Divide by the Coefficient of the Logarithm

To solve for \(\log x\), divide both sides of the equation by 10 (the coefficient of \(\log x\)):\[ \frac{D - 160}{10} = \log x \]

Step 3: Solve for x using Exponentiation

Since we have a logarithm base 10, rewrite the equation in exponential form to solve for \(x\). Recall that \( \log_{10}(x) = y \) implies \(10^y = x\).\[ x = 10^{\frac{D - 160}{10}} \]

Key Concepts

Isolation of Logarithmic Term

When solving equations involving logarithms, the first step is always to isolate the term that includes the logarithm. This simplifies the equation and allows you to solve subsequent steps more effectively.
In the problem provided, we start with the equation:

• \( D = 160 + 10 \log x \)

To isolate the logarithmic term, move the constant term \( 160 \) to the opposite side by subtracting it from both sides. This leads to:

• \( D - 160 = 10 \log x \)

This transformation is crucial as it sets the stage for the next steps. By isolating \( 10 \log x \), you establish a clear equation that directly correlates to the variable you wish to solve, \( x \). Remember to always handle algebraic expressions carefully when shifting terms around the equation in order to maintain equivalency.

Exponentiation to Solve Logarithms

After successfully isolating the logarithmic term, the next step in solving the equation involves exponentiating. Exponentiation is a powerful technique used to "undo" the logarithm, especially when the logarithm is in its standard base form.
Consider our equation post-isolation:

• \( \frac{D - 160}{10} = \log x \)

The equation above has \( \log x \), which can be rewritten using the property \( \log_{10}(x) = y \) implies \(10^y = x\). We apply this property to solve for \(x\) by converting the logarithmic equation into its exponential form:

• \( x = 10^{\frac{D - 160}{10}} \)

This process effectively removes the logarithmic function, thus allowing you to solve for \(x\) directly.
Exponentiation is effective because it leverages the inverse relationship between logarithms and exponents, allowing the transition from logarithmic form back to the arithmetic realm.

Manipulating Algebraic Expressions

Handling algebraic expressions carefully is key to successfully solving equations like the one in this problem. This involves performing operations that adhere to arithmetic rules and maintain equivalence throughout the equation.
To manipulate the expression and isolate \(\log x\), each operation needs to be applied equally to both sides of the equation.
First, you subtract \(160\) from both sides:

• \( D - 160 = 10 \log x \)

Next, divide every term by \(10\) to isolate \(\log x\):

• \( \frac{D - 160}{10} = \log x \)

These steps highlight the importance of balance. By performing identical operations on both sides, you ensure the equation remains balanced and true.
Keeping these principles in mind, complex problems can be simplified, paving the way to accurate solutions.