Problem 68 - Intermediate Algebra

Question

Solve for \(x .\) See Example 3. $$ \log _{x} \frac{1}{1,000}=-\frac{3}{2} $$

Step-by-Step Solution

Answer

The solution is \( x = 100 \).

Step 1: Understand the logarithm form

The equation \( \log_{x} \frac{1}{1,000} = -\frac{3}{2} \) tells you that the base \( x \) raised to the power \( -\frac{3}{2} \) results in \( \frac{1}{1,000} \). In logarithmic form, the general expression is \( y = \log_{b}(a) \) which translates to \( b^y = a \).

Step 2: Translate to exponential form

Convert the logarithmic equation into its exponential form: \( x^{-\frac{3}{2}} = \frac{1}{1,000} \). This means the variable \( x \) raised to the power \( -\frac{3}{2} \) equals \( \frac{1}{1,000} \).

Step 3: Simplify the right-hand side

Recognize that \( \frac{1}{1,000} \) can be rewritten using powers of 10: \( \frac{1}{1,000} = 10^{-3} \). This gives the equation \( x^{-\frac{3}{2}} = 10^{-3} \).

Step 4: Equate the exponents of the bases

The bases are now equal, so equate the exponents: \( -\frac{3}{2} \log_{10}(x) = -3 \). Divide both sides by \(-\frac{3}{2}\) to solve for \( \log_{10}(x) \). This results in \( \log_{10}(x) = 2 \).

Step 5: Solve for x

Convert \( \log_{10}(x) = 2 \) back into exponential form: \( x = 10^{2} \). Calculate \( 10^{2} \) which equals 100. Thus, \( x = 100 \).

Key Concepts

Exponential Form

In mathematics, the exponential form is a way to express the power to which a base number is raised to achieve a certain value. For example, if you have an equation like \( b^y = a \), you would read this as "base \( b \) raised to the power \( y \) equals \( a \)."
This concept is crucial in understanding how logarithms work. In reference to the provided exercise, when you have \( \,\log_{x} \frac{1}{1,000} = -\frac{3}{2} \,\), it helps to turn this into an exponential form to resolve for the base \( x \), converting it to \( x^{-\frac{3}{2}} = \frac{1}{1,000} \).
In this notation, \( x \) is the base, \(-\frac{3}{2}\) is the exponent, and \(\frac{1}{1,000}\) is the resulting value. This is helpful as operations in exponential form tend to be more straightforward, facilitating easier manipulation and solution discovery.

Logarithmic Form

The logarithmic form enables us to handle exponential equations by using logarithms. In standard terms, \( y = \log_{b}(a) \) can be translated to the exponential form \( b^y = a \). What this essentially means is that 'the logarithm of \( a \) to the base \( b \) is \( y \).' For instance, if \( \log_{10}(100) = 2 \), this is equivalent in exponential form as \( 10^2 = 100 \).
In the exercise from the original document, the logarithmic form \( \,\log_{x} \frac{1}{1,000} = -\frac{3}{2} \,\) denotes that the base \( x \) raised to the power of \(-\frac{3}{2}\) represents \(\frac{1}{1,000}\). Converting from logarithmic to exponential form simplifies solving for unknown bases, utilizing known properties of logarithms and exponents efficiently.

Solving for x

Solving for \( x \) involves rewriting the equations in a form that isolates \( x \) on one side. In our exercise, after translating the given logarithmic equation to \( x^{-\frac{3}{2}} = \frac{1}{1,000} \), we simplified \( \frac{1}{1,000} \) as \( 10^{-3} \), aligning the bases.
Next, by equating the exponents, we develop a new equation: \( -\frac{3}{2} \log_{10}(x) = -3 \). Solving this, we divide both sides by \(-\frac{3}{2}\) giving us \( \log_{10}(x) = 2 \). Converting back to exponential form, we find \( x = 10^2 \). Hence, solving for \( x \) reveals \( x = 100 \).
This process showcases how translating between logarithmic and exponential forms is a pivotal step in isolating the variable and attaining the solution.