Problem 76 - Calculus for Biology and Medicine

Question

Which real number \(x\) satisfies (a) \(\log _{1 / 2} x=-4\) ? (b) \(\log _{1 / 4} x=2 ?\) (c) \(\log _{5} x=3\) ?

Step-by-Step Solution

Answer

(a) 16, (b) 1/16, (c) 125.

Step 1: Understanding Logarithmic Equations

The equations provided in parts (a), (b), and (c) are all in logarithmic form. To find the value of the variable involved, each logarithmic equation can be rewritten in exponential form.

Step 2: Rewriting (a) in Exponential Form

Given \(\log_{1/2} x = -4\), this can be rewritten as an exponential equation: \((1/2)^{-4} = x\).

Step 3: Calculating (a)

Solving the exponential equation \((1/2)^{-4} = x\), we find \( (1/2)^{-4} = (2)^4 = 16 \). Hence, \( x = 16 \).

Step 4: Rewriting (b) in Exponential Form

For \(\log_{1/4} x = 2\), rewrite it as an exponential equation: \((1/4)^2 = x\).

Step 5: Calculating (b)

Solve \((1/4)^2 = x\) to find \( (1/4)^2 = 1/16 \), so \( x = 1/16 \).

Step 6: Rewriting (c) in Exponential Form

Given \(\log_5 x = 3\), rewrite it as \(5^3 = x\).

Step 7: Calculating (c)

Solve \(5^3 = x\) to find \( 5^3 = 125 \), so \( x = 125 \).

Key Concepts

Exponential Form

The exponential form is a crucial concept when working with logarithmic equations. By converting a logarithmic equation into exponential form, you can often solve the equation more easily. For instance, if you have a logarithmic equation like \(\log_b a = c\), you can rewrite this in exponential form as \(b^c = a\). In this transformation:

• \(b\) is the base of the logarithm, which becomes the base of the exponent.
• \(c\) is the logarithm of \(a\), representing the exponent.
• \(a\) is the result of raising \(b\) to the power of \(c\).

By rewriting logarithmic equations in this way, equations often become simpler and more intuitive to solve.

Solving Equations

Once you've rewritten a logarithmic equation into its exponential form, solving it becomes the next step. Let's look at this process with examples from the original exercise:

• For part (a), the equation \(\log_{1/2} x = -4\) becomes \( (1/2)^{-4} = x\). Solving this equation involves calculating \( (1/2)^{-4} \), which is \(16\).
• In part (b), \(\log_{1/4} x = 2\) is rewritten as \( (1/4)^2 = x\). Calculating \((1/4)^2\) gives \(1/16\).
• For part (c), \(\log_5 x = 3\) is rewritten as \(5^3 = x\). Solving gives \(5^3 = 125\).

This methodical approach shows that solving exponential equations often depends on precise calculation of powers and roots.

Exponent Laws

Understanding the laws of exponents is essential when dealing with exponential form or solving equations resulting from logarithmic problems. These laws describe how exponents behave under various operations, such as multiplication or division.

• Multiplying Powers with the Same Base: Add the exponents, \(a^m \times a^n = a^{m+n}\).
• Dividing Powers with the Same Base: Subtract the exponents, \(\frac{a^m}{a^n} = a^{m-n}\).
• Power of a Power: Multiply the exponents, \((a^m)^n = a^{m \cdot n}\).
• Negative Exponents: Represent the reciprocal, \(a^{-m} = \frac{1}{a^m}\).

These properties help simplify expressions and solve exponential equations, making them more manageable within the context of logarithmic solutions. For example, in the original exercise, negative exponents are employed to solve \((1/2)^{-4} = x\), converting it to a more familiar form.