Evaluate Logarithmic Expressions - Algebra-2

1. Fundamental Concepts

Definition of a Logarithmic Expression:
The logarithmic expression $\log_b x$ represents "the exponent to which the base $b$ must be raised to obtain the argument $x$". In other words, if $\log_b x = y$, it is equivalent to $b^y = x$ (where $b>0$, $b\neq1$, and $x>0$). The essence of evaluating a logarithmic expression is to find the value of the exponent $y$ that satisfies the above equivalence relation.
Domain and Range:
The argument $x$ must be greater than 0, and the base $b$ must be greater than 0 and not equal to 1. Otherwise, the logarithmic expression is meaningless and cannot be evaluated;

The result $y$ (i.e., the exponent) of the logarithmic expression can be any real number, such as an integer, a fraction, or a negative number, depending on the values of $b$ and $x$.

2. Key Concepts

Core of Equivalence Conversion:
The core of evaluating $\log_b x$ is to convert the logarithmic problem into an exponential problem. This is done by letting $\log_b x = y$, which transforms the expression into $b^y = x$, and then solving for $y$. This serves as the bridge connecting logarithms and exponents, and is the fundamental approach to evaluating logarithmic expressions.
Evaluation Characteristics of Special Logarithms:
Common logarithm $\lg x$ (i.e., $\log_{10} x$): When evaluating, we need to find the exponent to which 10 must be raised to get $x$. For example, to evaluate $\lg 100$, we find $y$ in $10^y = 100$;

Logarithms with special values: When $x = b$, $\log_b b = 1$ (since $b^1 = b$); when $x = 1$, $\log_b 1 = 0$ (since $b^0 = 1$). These special cases can be evaluated quickly using their inherent properties.
Power Relationship Between Base and Argument:
If the argument $x$ is an integer power, fractional power, or negative power of the base $b$, we can express $x$ as a power of $b$ (i.e., $x = b^k$, where $k$ is a real number). This allows us to directly conclude that $\log_b x = k$, simplifying the evaluation process.

3. Examples

Easy

Example 1: Evaluate $\log_4 64$

Step 1: Let $\log_4 64 = y$. According to the definition, convert it to the exponential equation $4^y = 64$;

Step 2: Analysis: Since $4^3 = 4\times4\times4 = 64$, we have $y = 3$;

Result: $\log_4 64 = 3$.

Example 2: Evaluate $\log_6 36$

Step 1: Let $\log_6 36 = y$, and convert it to $6^y = 36$;

Step 2: Analysis: $6^2 = 6\times6 = 36$, so $y = 2$;

Result: $\log_6 36 = 2$.

Medium

Example 1: Evaluate $\log_5 \frac{1}{5}$

Step 1: Let $\log_5 \frac{1}{5} = y$ , and convert it to $5^y = \frac{1}{5}$ ;

Step 2: Analysis: Express $\frac{1}{5}$ as a power of 5, i.e., $\frac{1}{5} = 5^{-1}$ . Thus, $y = -1$ ;

Result: $\log_5 \frac{1}{5} = -1$ .

Example 2: Evaluate $\log_{10} 0.001$

Step 1: Let $\log_{10} 0.001 = y$, and convert it to $10^y = 0.001$;

Step 2: Analysis: $0.001 = \frac{1}{1000} = 10^{-3}$, so $y = -3$;

Result: $\log_{10} 0.001 = -3$.

Hard

Example 1: Evaluate $\log_{2^3} 32$ (i.e., $\log_8 32$)

Step 1: Let $\log_8 32 = y$, and convert it to the exponential equation $8^y = 32$;

Step 2: Analysis: First, express the base and the argument as powers of the same prime number.

Since $8 = 2^3$ and $32 = 2^5$, the equation becomes $(2^3)^y = 2^5$;

Using the power of a power property $(a^m)^n = a^{mn}$, simplify it to $2^{3y} = 2^5$;

Since the bases are the same and not equal to 1, their exponents must be equal.

Thus, $3y = 5$, and solving for $y$ gives $y = \frac{5}{3}$;

Result: $\log_8 32 = \frac{5}{3}$.

Example 2: Evaluate $\log_{\sqrt{5}} 125$

Step 1: Let $\log_{\sqrt{5}} 125 = y$, and convert it to $(\sqrt{5})^y = 125$;

Step 2: Analysis: Unify the base to 5.

Since $\sqrt{5} = 5^{\frac{1}{2}}$ and $125 = 5^3$, the equation becomes $(5^{\frac{1}{2}})^y = 5^3$;

Using the power of a power property, we get $5^{\frac{y}{2}} = 5^3$.

Therefore, $\frac{y}{2} = 3$, and solving for $y$ gives $y = 6$;

Result: $\log_{\sqrt{5}} 125 = 6$.

4. Problem-Solving Techniques

"Variable Setting and Conversion" Method (Core Technique):
1. Let the logarithmic expression to be evaluated be $\log_b x = y$ (where $y$ is the unknown to be found);
2. Based on the equivalence relation between logarithms and exponents, convert it into the exponential equation $b^y = x$;

3. Solve for $y$ in the exponential equation, which is the result of the logarithmic expression.

Application Scenario: Evaluation of all logarithmic expressions, especially basic problems. It is the most universal problem-solving approach.
"Base Unification" Method:
1. When both the base b and the argument x of the logarithm are powers of a certain prime number (e.g., 2, 3, 5), first express both b and x as powers of that prime number (e.g., , , where a is a prime number);
2. Use the power of a power property to convert the exponential equation into ;
3. Since the exponents of equal powers with the same non-1 base are equal, we get and solve for y.
- Application Scenario: Evaluation of complex logarithms where the base and the argument share a common prime factor.
"Special Value Memory" Method:
1. Keep in mind the special value properties of logarithms: $\log_b b = 1$ (the logarithm of any number with itself as the base is 1) and $\log_b 1 = 0$ (the logarithm of 1 with any valid base is 0);

2. When the argument is equal to the base or the argument is 1, directly use the above properties to get the result without complex calculations.

Application Scenario: Evaluation of logarithms involving special arguments (the base itself, 1), such as $\log_7 7$, $\log_{0.5} 1$.
"Verification and Backtracking" Method:
1. After solving for the value of y, calculate and verify if it equals the argument x;
2. If , then y is the correct result; if not, check for errors in steps such as base conversion and power calculation during the problem-solving process.
- Application Scenario: Result verification for all logarithmic evaluations. It is particularly useful for ensuring accuracy when dealing with error-prone cases such as fractional exponents and negative exponents.