Problem 67 - Intermediate Algebra

Question

Solve for \(x .\) See Example 3. $$ \log _{100} x=\frac{3}{2} $$

Step-by-Step Solution

Answer

The solution is \( x = 1000 \).

Step 1: Understand the Problem

We need to solve for \( x \) where \( \log_{100} x = \frac{3}{2} \). This involves finding the value of \( x \) such that 100 raised to the power of \( \frac{3}{2} \) equals \( x \).

Step 2: Rewrite the Logarithmic Equation as an Exponential Equation

The logarithmic equation \( \log_{100} x = \frac{3}{2} \) can be rewritten in exponential form as \( 100^{\frac{3}{2}} = x \). This is because a logarithm \( \log_b a = c \) is equivalent to the exponential form \( b^c = a \).

Step 3: Calculate the Exponent

We need to calculate \( 100^{\frac{3}{2}} \). Recognize that \( \frac{3}{2} \) is \( 1 + \frac{1}{2} \), so \( 100^{\frac{3}{2}} = 100^1 \times 100^{\frac{1}{2}} \).

Step 4: Simplify the Expression

Calculate \( 100^{1} \), which is 100, and \( 100^{\frac{1}{2}} \). Since \( 100^{\frac{1}{2}} \) is the square root of 100, it equals 10. Thus, \( 100^{\frac{3}{2}} = 100 \times 10 = 1000 \).

Step 5: Conclude the Solution

The value of \( x \) that satisfies the equation \( \log_{100} x = \frac{3}{2} \) is \( x = 1000 \).

Key Concepts

Exponential Equations

An exponential equation involves expressions where variables appear as exponents. These are crucial in various fields, ranging from compound interest calculations to population growth models.
Understanding exponential equations is foundational in math because they represent constant rates of growth or decay.

• Typically, you'll see something like \(a^x = b\) where \(x\) is the exponent to be solved for.
• Exponential equations can be rewritten using logarithms, which are inverse operations.
• In this exercise, the exponential form was \(100^{\frac{3}{2}} = x\), showing \(x\) as the base, 100, raised to the power \(\frac{3}{2}\).

By rewriting logarithms as exponentials, we can solve for unknowns by applying rules of exponents and operations like squaring or finding roots.

Solving for x

To solve for \(x\) in any equation, the main goal is to isolate \(x\) on one side of the equation.
This often requires transforming equations and using mathematical operations strategically.
When dealing with logarithmic or exponential equations, it's key to understand the properties of logarithms and exponents.

• For a logarithmic equation like \(\log_{b} x = c\), the general approach is to rewrite it in exponential form \(b^c = x\).
  This step makes solving for \(x\) straightforward.
• Mathematical operations, like taking the square root or cube root, are often needed as part of the simplification.
• In this specific example, after rewriting the logarithmic equation in its exponential form, solving for \(x\) required calculating \(100^{1} \times 100^{\frac{1}{2}}\), yielding \(x = 1000\).

Practicing these processes enhances algebraic skills and problem-solving capabilities.

Logarithmic to Exponential Form

Converting logarithmic expressions into exponential form is a vital skill in algebra. This conversion is particularly useful when solving logarithmic equations.
Basically, this process flips the equation between two formats functionally equivalent but algebraically advantageous in different situations.

• Logarithmically, \(\log_{b} a = c\) denotes that \(b\) raised to the power of \(c\) equals \(a\).
• By converting, you enable straightforward computation of \(x\) when given a logarithmic question: transform \(\log_{100} x = \frac{3}{2}\) to \(100^{\frac{3}{2}} = x\).
• This method allows solving by computation because exponentiation is simpler to automate and execute with basic operations.

Converting between these forms aids comprehension and efficiency when tackling algebraic challenges.