Problem 88 - Algebra and Trigonometry

Question

Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\)Write each expression in terms of \(A\) and \(C\). $$ \log _{b} \sqrt{\frac{3}{16}} $$

Step-by-Step Solution

Answer

\(\log_{b} \sqrt{3/16} = 0.5C - 2A\)

Step 1: Convert the expression

First, \(\log_{b}\sqrt{3/16}\) can be converted to \(\frac{1}{2} \log_{b}(3/16)\) using the property of logarithms, which is \(\log_{b} a^n = n\log_{b} a\).

Step 2: Split the logarithm

The expression can be further split into \(\frac{1}{2} (\log_{b}3 - \log_{b}16)\) using another property of logarithms, \(\log_{b}(a/b) = \log_{b}a - \log_{b}b\).

Step 3: Simplify logarithms

Then the logarithm expression can be simplified to \(\frac{1}{2} (C - 4A)\), because \(\log_{b}3\) equals C, and \(\log_{b}16\) can be further simplified to \(4\log_{b}2 = 4A\), using the given \(\log_{b}2 = A\).

Step 4: Final simplification

Finally, simply multiply through by the 1/2 to get \(0.5C - 2A\).

Key Concepts

Logarithmic Functions

A logarithmic function is the inverse operation of an exponential function and it's fundamental in understanding various phenomena in science and mathematics. It is represented as \(y = \log_{b}x\), where \(b\) is the base of the logarithm, \(x\) is the argument, and \(y\) is the result or logarithm. The base \(b\) is a positive real number, and for any \(x > 0\), there exists a unique \(y\) such that \(b^y = x\). This can be markedly useful, as logarithmic functions help transform multiplicative relationships into additive ones, simplifying complex calculations.

Additionally, one of the most commonly used logarithmic functions is the natural logarithm, which has the base \(e\), an irrational constant approximately equal to 2.71828. It's denoted by \(\ln(x)\) rather than \(\log_{e}(x)\) for convenience. Logarithmic functions are continuous, one-to-one, and are defined for all positive real numbers when considering real number inputs.

Logarithmic Expressions

Logarithmic expressions represent the logarithm of a number or algebraic expression. When working with these expressions, it's crucial to be conversant with certain properties of logarithms that make manipulation and simplification possible.

Key Properties of Logarithmic Expressions:

• \textbf{Product Property:} \(\log_{b}(xy) = \log_{b}(x) + \log_{b}(y)\), which allows the logarithm of a product to be expressed as the sum of the logarithms.
• \textbf{Quotient Property:} \(\log_{b}(\frac{x}{y}) = \log_{b}(x) - \log_{b}(y)\), expressing the logarithm of a quotient as the difference of logarithms.
• \textbf{Power Property:} \(\log_{b}(x^p) = p\log_{b}(x)\), where a logarithm with an exponent can be 'brought down' and multiplied in front of the logarithm.
• \textbf{Change of Base Formula:} For \(b, a > 0\) and \(b eq 1\), \(\log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)}\) for any positive \(c\), \(c eq 1\).
• \textbf{Logarithm of One:} \(\log_{b}(1) = 0\) for any positive base \(b\) not equal to one.
• \textbf{Logarithm of the Base:} \(\log_{b}(b) = 1\), suggesting any base logged with itself equals one.

Understanding these properties lays down the foundation for solving more complex logarithmic expressions with ease.

Solving Logarithmic Equations

Solving logarithmic equations involves finding the value of the variable that makes the equation true. The approach usually requires one to be familiar with logarithmic properties, which allows for the manipulation and simplification of logarithmic terms. Here’s a concise step-by-step overview of the process:

Approach for Solving Logarithmic Equations:

• Make sure the equation is in logarithmic form, and simplify using properties of logarithms.
• Isolate the logarithmic term if possible.
• If there are logarithms on both sides of the equation with the same base, typically, they can be set equal to each other, omitting the logarithmic part (as \(\log_{b}(x) = \log_{b}(y)\) implies \(x = y\)).
• Convert logarithmic equations into exponential form to solve for the variable — using the definition that \(\log_{b}(x) = y\) is equivalent to \(b^{y} = x\).
• Check your answers by substituting them back into the original equation. This is an essential step as sometimes the logarithmic properties used might introduce extraneous solutions.

When comfortable with these steps, solving logarithmic equations, such as the one in the original problem, becomes a systemized process.