Problem 147 - Algebra and Trigonometry

Question

Check each proposed solution by direct substitution or with a graphing utility. $$ (\log x)(2 \log x+1)=6 $$

Step-by-Step Solution

Answer

The solutions to the equation \( (\log x)(2 \log x + 1) = 6 \) are \( x = 10 \) and \( x = 0.001 \)

Step 1: Simplify the equation

Firstly, the logarithmic equation \( \log x(2 \log x + 1) = 6 \) can be simplified by using the property of logarithms \( \log a^n = n \cdot \log a \) as follows: \( 2 (\log x)^2 + \log x - 6 = 0 \). Now, the equation looks like a quadratic equation.

Step 2: Solve the quadratic equation

Solving the quadratic equation for \( \log x \). This would give two possible results for \( \log x \). Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for a quadratic equation \( ax^2 + bx + c \). Our variables will be \( a = 2, b = 1 \) and \( c = -6 \). So, \( \log x = \frac{-1 \pm \sqrt{1 + 4 \cdot 2 \cdot 6}}{2 \cdot 2} \). Calculating this gives two values for \( \log x \), those are \( \log x = 1 \) and \( \log x = -3 \)

Step 3: Solve the solutions for x

We have two possible logarithmic solutions, so we solve for \( x \) by converting \( \log x = 1 \) and \( \log x = -3 \) into exponential form, getting \( x=10^1 \) and \( x=10^{-3} \) respectively.

Step 4: Check the solutions

In our original equation, we originally have \( \log x \), so x cannot be negative or zero. The possible solutions are \( x = 10 \) and \( x = 0.001 \). The solutions should be checked by substituting them back into the original equation to ensure they hold true.

Key Concepts

Logarithms

Logarithms are a fascinating mathematical concept. Imagine wanting to know what power you need to raise a number, called the base, to reach another specific number. Logarithms help with exactly that! The equation \(\log_b(a) = c\) means that \(\) base \( b\) raised to the power of \(\) \(c\) equals \( a\).
The most common bases are 10 (common logarithm) and \(e\) (natural logarithm).

• Use the property \(\log(xy) = \log x + \log y\) for multiplication.
• Transform powers with \(\log(x^n) = n \cdot \log x\).

When solving equations like \((\log{x})(2\log{x} + 1) = 6\), simplifying using these properties helps transform it into a more manageable quadratic equation.

Exponential Form

Exponential form is like translating from the "language" of logarithms into something more understandable.
After solving for \(\log x\) in the logarithmic equation, we get solutions like \(\log x = 1\) and \(\log x = -3\).
To convert these back into a number \(x\), we use the exponential form. This entails rewriting the expressions as \(x = 10^1\) and \(x = 10^{-3}\).

• Any number raised to 1 is itself, so \(10^1 = 10\).
• Raising to a negative power gives a fraction, so \(10^{-3} = 0.001\).

Understanding this conversion is crucial. It links the idea of exponents with the power of logarithms, providing concrete solutions for \(x\).

Algebraic Substitution

Algebraic substitution is like trading parts of an equation for something easier to work with.
In our case, after simplifying the logarithmic equation to \(2 (\log x)^2 + \log x - 6 = 0\), it takes on the form of a quadratic equation.
This is where substitution shines!
Consider \(\log x\) as a variable, say \(y\), transforming the equation into \(2y^2 + y - 6 = 0\).

• Now, solve for \(y\) using the quadratic formula.
• Then revert by substituting back to find \(x\).

This method not only simplifies complex expressions but also makes solving them structured and organized, revealing hidden relationships between variables.

Graphing Utility

Graphing utilities are digital tools that visualize mathematical equations, bringing clarity to potentially confusing problems.
In exercises like ours, they show intersections where solutions lie.
By plotting the transformed quadratic equation \(2 (\log x)^2 + \log x - 6 = 0\), the graph displays solutions as points where it crosses the x-axis.

• These visual cues help verify algebraic solutions.
• They provide insight into the function's behavior.

Thus, using graphing utilities is like seeing the problem from a new angle. It enriches understanding by clearly showing where and why certain solutions work, especially in the context of complex equations.