Problem 68 - Precalculus with Calculus Previews

Question

Solve the given equation. $$ x^{\log _{10} x}=\frac{1000}{x^{2}} $$

Step-by-Step Solution

Answer

The solutions are \( x = 10 \) and \( x = 0.001 \).

Step 1: Rewrite the Equation Using Properties of Logarithms

The original equation is given as \[ x^{\log _{10} x} = \frac{1000}{x^{2}} \] To simplify, take the logarithm on both sides to leverage properties of exponents. This converts the power into a multiplication: \[ \log_{10}(x) \times \log_{10}(x) = \log_{10}(1000) - 2\log_{10}(x) \] Since \( \log_{10}(1000) = 3 \), the equation becomes: \[ (\log_{10}x)^2 + 2\log_{10}x - 3 = 0 \]

Step 2: Solve the Quadratic Equation

Using the quadratic formula \( ax^2 + bx + c = 0 \), the solutions are \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 2 \), and \( c = -3 \).Substituting these values, we find: \[ \log_{10}x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-3)}}{2 \times 1} = \frac{-2 \pm \sqrt{4 + 12}}{2} \] \[ \log_{10}x = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2} \] Now solve for two possible values: 1. \( \log_{10}x = \frac{2}{2} = 1 \) 2. \( \log_{10}x = \frac{-6}{2} = -3 \)

Step 3: Convert Logarithmic Solutions Back to Exponential Form

The two solutions for \( \log_{10}x \) are 1 and -3.1. If \( \log_{10}x = 1 \), then \( x = 10^1 = 10 \).2. If \( \log_{10}x = -3 \), then \( x = 10^{-3} = 0.001 \).

Key Concepts

Logarithmic Equations

Logarithmic equations involve unknowns contained within a logarithm. These equations are often solved by converting the logs into another form or simplifying using properties of logarithms.
To tackle these equations, first, understand how to use the properties of logs to isolate the variable.
For example, in the equation \( x^{\log _{10} x} = \frac{1000}{x^{2}} \), applying the logarithm to both sides helps convert the exponents involved into a form that can be easier to manage.
Essentially, you can take:

• The logarithm of both sides of the equation
• Use logarithms to transform multiplication and division into addition and subtraction, respectively
• Solve for the unknown variable once the equation is simplified

Understanding how to use these properties efficiently allows you to solve complex-looking equations without excessive complexity.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). In this equation, there is always a term with a variable squared.
Solving quadratic equations involves finding the values for which the equation is satisfied.
To solve quadratic equations, the quadratic formula is one of the most reliable and common methods.
The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• This formula allows you to find solutions for any quadratic equation

In the given exercise, after converting the logarithm into a quadratic form \( (\log_{10}x)^2 + 2\log_{10}x - 3 = 0 \), the quadratic formula was used to find values of \( \log_{10}x \). After finding these, converting them back to solve for \( x \) is straightforward.

Exponential Form

Exponential form involves expressing numbers using a base raised to a power. This form is especially handy when dealing with equations where the variable is an exponent.
Converting logarithmic solutions to exponential form is crucial for finding the actual solutions for \( x \).
For instance:

• If \( \log_{10}x = 1 \), then the exponential form is \( x = 10^1 = 10 \)
• If \( \log_{10}x = -3 \), then the exponential form is \( x = 10^{-3} = 0.001 \)

This transition from logarithmic to exponential is crucial for interpreting your solutions.
By transforming the problem back into an exponential form, you're able to determine concrete numeric values for the solutions, \( x \).
This conversion showcases how exponential growth or decay is represented in practical scenarios.

Properties of Logarithms

The properties of logarithms are essential tools that help in simplifying and solving logarithmic equations. These properties allow us to manipulate logs into a more solvable form.
Some key properties include:

• \( \log_b(mn) = \log_b(m) + \log_b(n) \)
• \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \)
• \( \log_b(m^n) = n\log_b(m) \)

In our equation exercise, using these properties allowed us to simplify the expression by turning multiplication into addition or division into subtraction.
For example, we utilized \( \log_{10}(x^2) = 2\log_{10}(x) \) to simplify terms. Understanding and applying these properties means you can solve complex logarithmic equations more effectively.
They form the backbone of any manipulation or simplification in problems involving logarithms.