Problem 81 - Calculus for Biology and Medicine

Question

Simplify the following expressions: (a) \(2^{5 \log _{2} x}\) (b) \(3^{4 \log _{3} x}\) (c) \(5^{5 \log _{1 / 5} x}\) (d) \(4^{-2 \log _{2} x}\) (e) \(2^{3 \log _{1 / 2} x}\) (f) \(4^{-\log _{1 / 2} x}\)

Step-by-Step Solution

Answer

(a) \(x^5\), (b) \(x^4\), (c) \(x^{-5}\), (d) \(x^{-4}\), (e) \(x^{-3}\), (f) \(x^2\).

Step 1: Recognize the Logarithmic Identity

We start by using the identity for logs and exponents: if \( a^{b \log_a c} \), then it simplifies to \( c^b \). This identity will help us simplify each expression.

Step 2: Simplify Expression (a)

Expression (a) is \(2^{5 \log_{2} x}\). Using the identity, we find that it simplifies to \(x^5\).

Step 3: Simplify Expression (b)

Expression (b) is \(3^{4 \log_{3} x}\). Applying the identity gives us \(x^4\).

Step 4: Simplify Expression (c)

Expression (c) is \(5^{5 \log_{1 / 5} x}\). Recall that \(\log_{1 / 5} x = -\log_{5} x\). Therefore, it becomes \(5^{5(-\log_{5} x)} = 5^{-5 \log_{5} x}\), which simplifies to \(x^{-5}\).

Step 5: Simplify Expression (d)

Expression (d) is \(4^{-2 \log_{2} x}\). Notice that \(4 = 2^2\), so \(4^{-2 \log_{2} x} = (2^2)^{-2 \log_{2} x} = 2^{-4 \log_{2} x}\). Simplifying gives \(x^{-4}\).

Step 6: Simplify Expression (e)

Expression (e) is \(2^{3 \log_{1 / 2} x}\). Since \(\log_{1 / 2} x = -\log_{2} x\), it becomes \(2^{3(-\log_{2} x)} = 2^{-3 \log_{2} x} = x^{-3}\).

Step 7: Simplify Expression (f)

Expression (f) is \(4^{-\log_{1 / 2} x}\). We see that \(4 = 2^2\) and \(\log_{1 / 2} x = -\log_{2} x\). Thus, \(4^{-\log_{1 / 2} x} = (2^2)^{-(-\log_{2} x)} = 2^{2 \log_{2} x}\), which simplifies to \(x^2\).

Key Concepts

Understanding Exponents

Exponents are a shorthand way of representing repeated multiplication of a number by itself. The expression \(a^b\) means you multiply the base \(a\) by itself \(b\) times. This concept is fundamental in mathematics because it allows us to describe large numbers in a compact form. For example, \(2^3\) is \(2 \times 2 \times 2 = 8\).
Exponents have several rules that make calculations easier:

• Multiplying with the Same Base: \(a^m \cdot a^n = a^{m+n}\)
• Dividing with the Same Base: \(\frac{a^m}{a^n} = a^{m-n}\)
• Power of a Power: \((a^m)^n = a^{m \cdot n}\)
• Zero Exponent: \(a^0 = 1\) (for \(a eq 0\))
• Negative Exponent: \(a^{-n} = \frac{1}{a^n}\)

Understanding these rules helps simplify expressions and solve equations involving exponents quickly and effectively.

The Nature of Logarithms

Logarithms can be thought of as the inverse operation of exponentiation. If \(a^b = c\), then \(\log_a c = b\). This means a logarithm answers the question: "To what power must the base be raised to produce a given number?"
An important property of logarithms is that they simplify exponentiation processes. For instance:

• Product to a Sum \(\log_a(xy) = \log_a x + \log_a y\)
• Quotient to a Difference \(\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y\)
• Power Rule \(\log_a(x^b) = b \cdot \log_a x\)
• Change of Base \(\log_b x = \frac{\log_k x}{\log_k b}\)

The exercises you are encountering use a specific identity: \(a^{b \log_a c} = c^b\). This identity is crucial in simplifying expressions involving exponents with logarithmic exponents, as it effectively "cancels out" the logarithmic function.

Simplification of Expressions

Simplifying mathematical expressions means reducing them to their simplest form. With exponents and logarithms, simplification often involves applying the rules and properties of these operations.
The key to simplification in the given expressions is recognizing patterns. You can use:

• The identity \(a^{b \log_a c} = c^b\) to eliminate complex exponentials.
• Conversion of bases: Express a number like \(4\) as \((2^2)\) to simplify powers.
• Understanding negative logarithms: \(\log_{1/a} x = -\log_a x\).

These steps factor into how we approach simplifying expressions in examples such as:

• Turning \(2^{5 \log_2 x}\) into \(x^5\), by directly applying the identity.
• Rewriting \(4^{-2 \log_2 x}\) as \(x^{-4}\), by using both conversion of base and applying properties of exponents.

Mastering these techniques ensures you can handle a wide range of logarithmic and exponential problems efficiently.