Problem 9 - College Algebra and Calculus: An Applied Approach

Question

Solve for \(x\).\(\log _{10} x=-1\)

Step-by-Step Solution

Answer

The solution to the equation \(\log _{10} x = -1\) is \(x = \frac{1}{10}\).

Step 1: Rewrite in exponential form

Convert the given logarithmic equation \(\log _{10} x = -1\) to its equivalent exponential form. Recall from the properties of logarithms that \(\log _{b} a = c\) is equivalent to \(b^n = a\). Here \(b\) is the base, \(a\) is the argument of the logarithm, and \(c\) is the value of the logarithm. Therefore, we can rewrite \(\log _{10} x = -1\) as \(10^{-1} = x\).

Step 2: Simplify Expression

Simplify \(10^{-1}\) to calculate the value of \(x\). Remember that any number raised to the power of -1 is equal to its reciprocal, thus, \(10^{-1} = \frac{1}{10}\).

Step 3: Solve for x

So, by equating the two, we get \(x = \frac{1}{10}\) as the solution to the given logarithmic equation.

Key Concepts

Exponential Form

Logarithmic equations, like the one provided in the exercise, are closely tied to the concept of exponential form.
To understand this relationship, we need to remember that logarithms are essentially a different way to express exponents.
For instance, if you have an equation like \( \log_b a = c \), it can be translated into exponential form as \( b^c = a \).
This conversion is powerful because it allows us to switch between logarithmic and exponential expressions effortlessly.
When we apply this to our original problem, \( \log_{10} x = -1 \), we convert it to exponential form by recognizing that the base is 10, the argument is \( x \), and the value is -1.
This translates to \( 10^{-1} = x \), giving us a clearer path to solving for \( x \).
This step simplifies the problem and highlights the versatility of understanding both forms.

Properties of Logarithms

The properties of logarithms are tools which simplify and solve equations efficiently. They are especially useful for rewriting logarithmic expressions and solving equations.
One such property is the conversion between logarithmic and exponential forms as discussed earlier.
Another useful property is understanding how to manipulate logarithms with different bases and exponents.
For example, \( \log_b b^n = n \), which means any base raised to the logarithm of the same base equals the argument.
This could help in unraveling more complex logarithmic equations.
Moreover, remember that the logarithm of 1 with any base is always 0, \( \log_b 1 = 0 \).
This helps to simplify expressions where you might find a log term centered around 1.These properties aid us in transitioning between different parts of an equation, making solutions more apparent and logical.

Solving Logarithmic Equations

Solving logarithmic equations involves converting them into a form you can simplify. Typically, this means converting them into exponential form.
Then, you carefully isolate \( x \) to solve for its value.
Here’s how:

• Start by expressing the logarithm as an exponential equation, as we did in our example: convert \( \log_{10} x = -1 \) to \( 10^{-1} = x \).
• Simplify the expression to find the numerical value: \( 10^{-1} \) simplifies to \( \frac{1}{10} \).
• Finally, set this equal to \( x \) to find your solution: \( x = \frac{1}{10} \).

Note that sometimes additional steps may be required if the equation is more complex, such as combining logarithmic terms or applying additional properties of logarithms.
By following these steps, you can systematically tackle and solve logarithmic equations effectively.