Question

Write each expression as a single logarithm. \(\log _{3} \sqrt{x}-\log _{3} x^{3}\)

Short answer

\text{\textbackslash log\textunderscore{3} x^{(-5/2)}

Step-by-step solution

Apply the logarithm property for roots and exponents

Recall that \(\text{\textbackslash log\textunderscore{a} b^{c} = c \text{\textbackslash log\textunderscore{a} b}}\). Also, \(\text{\textbackslash log\textunderscore{a} \text{\textbackslash sqrt{x}} = \text{\textbackslash log\textunderscore{a} x^{1/2}} = 1/2 \text{\textbackslash log\textunderscore{a} x}}\). Applying this, the given expression becomes \(1/2 \text{\textbackslash log\textunderscore{3} x - 3 \text{\textbackslash log\textunderscore{3} x}}\).

Combine the logarithms using properties

Since both terms involve \(\text{\textbackslash log\textunderscore{3} x}\), the expression \(1/2\text{\textbackslash log\textunderscore{3} x - 3\text{\textbackslash log\textunderscore{3} x}}\) can be combined as \((1/2 - 3)\text{\textbackslash log\textunderscore{3} x} = (-5/2)\text{\textbackslash log\textunderscore{3} x}\).

Rewriting the logarithm

Use the logarithm power rule to combine into a single logarithm: \(\text{\textbackslash log\textunderscore{a} b^{c} = c\text{\textbackslash log\textunderscore{a} b}}\). Therefore, \((-5/2) \text{\textbackslash log\textunderscore{3} x} = \text{\textbackslash log\textunderscore{3} x^{(-5/2)}}\).

Key concepts

Logarithm Rules

Understanding logarithm rules is essential for solving many logarithmic expressions. Logarithms have several properties that simplify how we manipulate and combine them. These properties often mirror the rules for exponents but in a logarithmic form. The main rules to remember include:
- Product Rule: \(\text{log}_a (bc) = \text{log}_a b + \text{log}_a c\) - This tells us that the logarithm of a product is the sum of the logarithms.
- Quotient Rule: \(\text{log}_a \frac{b}{c} = \text{log}_a b - \text{log}_a c\) - This states that the logarithm of a quotient is the difference of the logarithms.
- Power Rule: \(\text{log}_a (b^c) = c \text{log}_a b\) - This allows us to move an exponent in front of the logarithm.
- Change of Base Rule: \(\text{log}_a b = \frac{\text{log}_c b}{\text{log}_c a}\) - This is useful for converting between different bases.
Leveraging these rules effectively can greatly simplify complex logarithmic calculations.

Exponents in Logarithms

Exponents play a crucial role in logarithmic expressions. The Power Rule is particularly useful in transformations involving exponents within logarithms. For example:
\(\text{log}_3 (x^{3}) = 3 \text{log}_3 x\).
A less obvious but similarly useful rule involves roots, which are simply exponents in disguise. For example:
\(\text{log}_3 \text{sqrt{x}} = \text{log}_3 (x^{1/2}) = \frac{1}{2} \text{log}_3 x\).
By understanding these rules, we can transform and simplify expressions effectively. When applied correctly, they turn a complex problem into a simple one. For instance, in the given problem, we applied this rule to get \(1/2 \text{log}_3 x - 3 \text{log}_3 x\).

Combining Logarithms

Combining logarithms involves using the logarithm rules to merge multiple logarithmic terms into one. This is especially useful in simplifying expressions. The given exercise utilizes the known rules of logarithms to combine terms. For instance:
\(1/2 \text{log}_3 x - 3 \text{log}_3 x\)
Here, both terms involve \(\text{log}_3 x\), allowing us to factor out \(\text{log}_3 x\). We then get:
\((1/2 - 3)\text{log}_3 x = (-5/2)\text{log}_3 x\).
Combining logarithms in this manner makes it easier to further simplify and conclude our problem by turning it into a single logarithmic term.

Logarithmic Expressions

Logarithmic expressions often need to be simplified using the basic properties of logarithms. By systematically applying the rules discussed, we reframe complex expressions into simpler forms. For instance, the final expression in the exercise is:
\((-5/2) \text{log}_3 x\)
We then apply the power rule to rewrite it as a single logarithm:
\(\text{log}_3 x^{(-5/2)}\).
This shows how logarithmic expressions can be condensed and simplified, turning a seemingly complex problem into a straightforward solution. Understanding how to perform these steps improves problem-solving efficiency and strengthens your grasp of logarithmic concepts.