Problem 76 - Calculus for Biology and Medicine

Question

Which real number \(x\) satisfies (a) \(\log _{2} x=-3\) ? (b) \(\log _{1 / 4} x=-\frac{1}{2} ?\) (c) \(\log _{3} x=0 ?\)

Step-by-Step Solution

Answer

(a) \(x = \frac{1}{8}\), (b) \(x = 2\), (c) \(x = 1\).

Step 1: Understanding Logarithms

The logarithmic equation \(\log_b a = c\) means \(b^c = a\). In other words, it indicates what power we need to raise \(b\) to get \(a\). We will use this property to find the value of \(x\) in each case.

Step 2: Solve Part (a)

The problem states \(\log _{2} x = -3\). According to the logarithmic property, this means \(2^{-3} = x\). Calculating further, we find: \[2^{-3} = \frac{1}{2^3} = \frac{1}{8}\] Thus, the value of \(x\) for part (a) is \(x = \frac{1}{8}\).

Step 3: Solve Part (b)

The equation given is \(\log _{1/4} x = -\frac{1}{2}\). This means \((1/4)^{-1/2} = x\). Rewriting \(1/4\) as \(4^{-1}\), we have: \[(4^{-1})^{-1/2} = x\]Using the power of a power property \((a^{m})^{n} = a^{m \cdot n}\), we find: \[4^{1/2} = x\]Calculating, we find that \(4^{1/2} = \sqrt{4} = 2\). Therefore, \(x = 2\) for part (b).

Step 4: Solve Part (c)

The equation is \(\log _{3} x = 0\), which means \(3^0 = x\). Knowing that any number to the power of zero is 1, we have: \[x = 1\].

Step 5: Conclusion

We've found the values of \(x\) for the given logarithmic equations: (a) \(x = \frac{1}{8}\)(b) \(x = 2\)(c) \(x = 1\).

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is essential when solving logarithmic equations. Logarithms transform multiplication into addition and exponents into simple multiplication. This simplification is key in various fields like engineering and science. Logarithms have some specific properties that can make complex calculations more manageable.

• The property \(\log_b (mn) = \log_b m + \log_b n\), which translates multiplication inside the log into addition outside it.
• The property \(\log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n\), which allows division inside the log to become subtraction outside it.
• The power property \((\log_b m^n = n \cdot \log_b m)\), which demonstrates that a power inside the log becomes a multiplication outside it.

These properties help simplify and rearrange equations, making them easier to solve. When you're attempting to solve an equation, identifying which properties can be applied is a crucial skill. This is how we can handle exponents and break down the components in a logarithmic expression effectively.

Solving Logarithmic Equations

When confronting logarithmic equations, you need to focus on rewriting these equations in a form that's manageable and easier to understand. Solving these requires a combination of knowledge about logarithms and algebraic manipulation. Start by identifying the equation in the form \(\log_b a = c\), which can be rewritten as \(b^c = a\) using the definition of a logarithm.

• For example, in equation \(\log_2 x = -3\), you switch forms to \(2^{-3} = x\). This approach simplifies the structure, allowing you to calculate the value of \(x\).
• In another case, if presented with \(\log_{1/4} x = -1/2\), you convert it to \( (1/4)^{-1/2} = x\), facilitating direct calculation.

The main objective in each situation is to use the properties of exponents to find the solution for \(x\). This often means simplifying terms and evaluating powers or roots. Incrementally breaking down each part of the problem helps to verify steps and achieve the correct solution while understanding the reasoning behind it.

Exponents and Logarithms

Exponents and logarithms have a very tight relationship; understanding one can greatly enhance your comprehension of the other. At the core, logarithms ask the question of "how many of one number do we multiply to get another?" Whereas exponents provide the answer to "what is the result of multiplying a number by itself a specified number of times?" These operations are inverses of each other.

• Converting between exponential and logarithmic forms is crucial. \(b^c = a\) is the exponential form, and \(\log_b a = c\) is the logarithmic form.
• Understanding the \(b^c = a\) relation not only helps solve equations but also to grasp more complex concepts in advanced mathematics.
• The rules, like any number to the power of zero equaling one, are directly applicable, as seen when \(3^0\) equates to \(1\).

To fully understand the interplay between these two mathematical concepts, practice converting forms and solving basic to intermediate equations. This will prepare you for tackling more complex problems and enhances your overall mathematical problem-solving skills.