Problem 34 - Algebra 2

Question

Solve each equation. Check your answers. $$ 2 \log x=-1 $$

Step-by-Step Solution

Answer

The equation has no solution.

Step 1: Convert To Exponential Form

The equation is currently in logarithmic form. To solve for \(x\), we need to convert this into its equivalent exponential form. This involves taking the base (\(10\), since it's a common log) to the power of the number on the other side of the equation (\(-1\)) and setting that equal to \(x\). Thus, the equation becomes \(10^{-1} = x\).

Step 2: Evaluate the Power

Next, calculate the numerical result of \(10^{-1}\). Remember that any number raised to the negative power is equivalent to the reciprocal of the number raised to that power. Therefore, \(10^{-1} = 1/10\). This gives us \(x = 1/10\).

Step 3: Check the Solution

We must check the solution we've found to make sure it doesn't require us to take the logarithm of a negative number or zero (which is not permitted). Substituting \(x = 1/10\) back into the original equation \(2 \log x = -1\), we get \(2 \log (1/10) = -1\). Evaluating the left side yields \(-2\), not \(-1\), which means this solution is incorrect. The original equation has no solution.

Key Concepts

exponential form

Converting a logarithmic equation to its exponential form is a crucial step in solving for the unknown variable. When you see an equation like \(2 \log x = -1\), it involves understanding that the logarithm here is in base 10, known as the common logarithm. To convert this into exponential form follow these steps:

• Identify the base, which is 10 in this case.
• The exponent is what the logarithm results in; here, it’s \(-1\).
• Set the equation with the base raised to this exponent: \( 10^{-1} = x \).

The transformation into exponential form helps simplify the equation into a single power statement, making it easier to solve for \( x \). This conversion essentially asks, "To what power must 10 be raised to result in our value of \( x \)?" Once in this form, evaluating it becomes more straightforward.

negative exponents

Negative exponents can be a bit confusing at first, but they follow a simple rule: A negative exponent indicates the reciprocal of the base raised to a positive exponent. This means that if you encounter something like \(10^{-1}\), here's what happens:

• The negative exponent signifies that you're working with a fraction.
• The base (10 here) should be moved to the denominator.
• Convert \(10^{-1}\) to \(\frac{1}{10}\).

In general, any number raised to a negative exponent, \(a^{-n}\), is equivalent to \(\frac{1}{a^n}\). Understanding how to interpret and simplify expressions with negative exponents unlocks a lot of mathematical doorways and helps solve logarithmic and exponential equations accurately.

reciprocal of numbers

The reciprocal of a number is defined simply as \( \frac{1}{\text{Number}} \). For example, the reciprocal of 10 is \( \frac{1}{10} \). Understanding reciprocals is crucial, especially when handling equations involving negative exponents. In our earlier problem, converting \(10^{-1}\) into \(\frac{1}{10}\) was possible due to the reciprocal concept.Key points about reciprocals include:

• Multiplying a number by its reciprocal always results in 1, i.e., \(a \times \frac{1}{a} = 1\).
• Reciprocals are particularly handy when simplifying division problems or negative exponent terms, which otherwise could seem complex at first glance.
• They enable us to handle fractions intuitively, turning division into multiplication.

Reciprocals simplify expressions and equations, making calculations neat and bringing clarity to potential confusion points.

checking solutions

Checking the solutions to an equation ensures accuracy and helps identify any possible errors. Not all initial solutions will satisfy the original equation, especially in the case of logarithmic equations where invalid solutions can emerge. From our example, we found an expression like \( 2 \log x = -1 \) seems to lead correctly to \( x = \frac{1}{10} \). However, checking the solution involves:

• Substituting \( x = \frac{1}{10} \) back into the original equation.
• Calculating the left side \( 2 \log (\frac{1}{10}) = -2 \), which does not equal \(-1\).

This demonstrates that the solution doesn't satisfy the equation, meaning no solution exists for the given equation within permissible real numbers for \( x \).Never skip the step of substituting back to verify because it confirms the validity of derived solutions and ensures understanding of constraints particular to logarithmic functions. This method of verification is crucial in honing problem-solving skills systematically.