Problem 12 - Precalculus

Question

Solve the equation analytically. $$ 10 \log \left(\frac{x}{10^{-12}}\right)=150 $$

Step-by-Step Solution

Answer

The solution is \(x = 1000\).

Step 1: Understanding the Given Equation

The given problem is an equation involving a logarithm: \(10 \log \left(\frac{x}{10^{-12}}\right) = 150\). This equation can be seen as a multiplication of the logarithmic expression by 10, which results in an equation set equal to 150.

Step 2: Isolate the Logarithmic Expression

First, divide both sides of the equation by 10 to isolate the logarithm: \(\log \left(\frac{x}{10^{-12}}\right) = 15\). This simplifies the equation to focus on solving for the logarithmic part.

Step 3: Convert from Logarithm to Exponential Form

Recall that if \(\log_{10}(A) = B\), then \(A = 10^B\). Using this, express the equation \(\log \left(\frac{x}{10^{-12}}\right) = 15\) as \(\frac{x}{10^{-12}} = 10^{15}\).

Step 4: Solve for \(x\)

Multiply both sides of the equation \(\frac{x}{10^{-12}} = 10^{15}\) by \(10^{-12}\) to solve for \(x\):\[x = 10^{15} \times 10^{-12}\].Combine the powers of 10: \(x = 10^{15-12} = 10^3\).

Step 5: Final Result

The solution to the equation is \(x = 10^3 = 1000\). Verify that this makes sense in the original equation.

Key Concepts

Logarithms

Logarithms are mathematical operations that are the inverse of exponentiation. Consider the expression

• If you have a log equation such as \(\log_{b}(A) = C\), the base \(b\) raised to the power of \(C\) equals \(A\). For instance, \(\log_{10}(100) = 2\) because \(10^2 = 100\).
• A logarithm answers the question: "To what power must we raise the base to obtain the given number?"
• They are especially useful in solving equations where the variable is an exponent.

This operation appears frequently in many branches of science and engineering, particularly in cases where exponential relationships take place.

Exponential Form

Exponential form refers to expressing numbers using a base and an exponent. When we deal with logarithms, often we will use exponential forms to simplify or solve equations.

• An equation \(\log_{10}(A) = B\) can be rewritten in exponential form as \(A = 10^B\).
• This conversion is crucial when you need to isolate the variable within a logarithm, allowing us to express a logarithmically simplified equation into something more straightforward.

In our exercise, converting \(\log(\frac{x}{10^{-12}}) = 15\) to exponential form involves recognizing \(\frac{x}{10^{-12}} = 10^{15}\). This form directly helps in solving the equation by making the variable more apparent.

Solving Equations

Solving equations involving logarithms starts with simplifying and sometimes changing their forms. For efficient solving:

• Isolate the logarithmic part. This typically involves algebraic manipulation, such as dividing or multiplying both sides of the equation by a constant.
• Once isolated, switch from logarithmic to exponential form.

This leads to a more straightforward algebraic expression where it's easier to solve for the unknown variable. As seen in our example, \(10 \log(\frac{x}{10^{-12}}) = 150\) was divided by 10 to focus on the logarithmic expression first.
Then, turning it into exponential form allowed us to resolve \(x\) directly.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to solve for an unknown variable. This is a fundamental skill when solving logarithmic equations:

• Start by eliminating coefficients that might multiply the logarithmic term, as seen by dividing both sides of the given equation by 10.
• Transition the equation into a form where you can use basic algebra operations, such as multiplication or division, to further simplify.
• In our exercise, after converting to exponential form, the next algebraic step was multiplying by \(10^{-12}\) to eliminate the fraction form, eventually leading to \(x = 10^3\).

Algebraic manipulation also includes combining or simplifying powers of 10 in our context, ensuring an accurate final result.