Exponential and Logarithmic Equations Practice Guide

Methodology for Solving Basic Exponential Equations

Section A of the practice set focuses on solving exponential equations in their basic form, where a base raised to an unknown power $x$ is equal to a constant. The prescribed method for solving these equations involves using the properties of logarithms to isolate the variable. Specifically, the instruction is to take the common logarithm ( $\log_{10}$ ) of both sides of the equation. This operation allows for the application of the power rule of logarithms, which states that $\log(b^x) = x \cdot \log(b)$ , thereby bringing the exponent down as a multiplier.

Once the equation is transformed into the form $x \cdot \log_{10}(\text{base}) = \log_{10}(\text{constant})$ , the variable $x$ is isolated by dividing the logarithm of the constant by the logarithm of the base: $x = \frac{\log_{10}(\text{constant})}{\log_{10}(\text{base})}$ . For the purposes of this practice set, all final values for $x$ must be rounded to exactly three decimal places.

Practice Problems: Basic Exponential Equations (Section A)

The following equations are to be solved using the $\log_{10}$ method described above, with rounding applied to the thousandths place:

(a) $5^x = 21$

(b) $8^x = 50$

(d) $9^x = 67$

(e) $6^x = 40$

(f) $11^x = 85$

(g) $3^x = 28$

Methodology for Solving Exponential Equations with Combined Exponents

Section B introduces equations where the left-hand side consists of a product of two exponential terms with the same base but different exponents. Before solving for the variable $x$ , these terms must be consolidated into a single exponential expression. This consolidation is achieved using the multiplication law of exponents, defined by the formula:

$a^m \cdot a^n = a^{m+n}$

By applying this rule, the exponents are summed together to create a single term with base $a$ . For example, a term like $5^x(5^{3x-1})$ would be rewritten as $5^{x + (3x - 1)}$ , which simplifies to $5^{4x-1}$ . After the left-hand side has been simplified into a single base raised to a power, the equation is solved following the same principles as basic exponential equations: taking the logarithm of both sides, applying the power rule, and performing algebraic division and addition/subtraction to isolate $x$ .

Practice Problems: Combined Exponents (Section B)

Solve for $x$ by first combining the exponents on the left-hand side of each equation:

(a) $5^x(5^{3x-1}) = 42$

(b) $2^x(2^{x+2}) = 19$

(d) $3^{x+1}(3^x) = 28$

(e) $7^x(7^{2x-1}) = 61$

(f) $2^{2x}(2^{x+1}) = 33$

(g) $6^x(6^{x+3}) = 75$

Methodology for Solving Logarithmic Equations

Section C covers logarithmic equations involving common logarithms ( $\log_{10}$ ). The strategy for these problems involves three distinct steps. First, if multiple log terms appear on one side of the equation separated by subtraction, they must be combined into a single logarithm using the Quotient Rule:

$\log(A) - \log(B) = \log\left(\frac{A}{B}\right)$

Second, once the equation is in the form $\log_{10}(\text{expression}) = \text{number}$ , it must be rewritten in its equivalent exponential form. Because these are common logarithms, the base is 10. The conversion follows the standard definition of logarithms: if $\log_{10}(y) = x$ , then $10^x = y$ . Third, the resulting linear or algebraic equation is solved for $x$ .

A critical requirement for logarithmic equations is the verification of solutions. Since the domain of a logarithmic function is restricted to positive real numbers, the solution for $x$ must be checked against the original log arguments (e.g., $(5x - 2)$ or $(x + 3)$ ). Any solution that results in an argument being zero or negative is considered extraneous and must be rejected.

Practice Problems: Logarithmic Equations (Section C)

The following equations require combining the logs, converting to base 10 exponential form, and validating the final solution:

(a) $\log_{10}(5x - 2) - \log_{10}(x + 3) = 1$

(b) $\log_{10}(8x + 1) - \log_{10}(x - 2) = 2$

(d) $\log_{10}(9x + 2) - \log_{10}(x + 6) = 1$

(e) $\log_{10}(4x - 1) - \log_{10}(x + 7) = 1$

(f) $\log_{10}(10x - 3) - \log_{10}(x + 4) = 2$

(g) $\log_{10}(3x + 5) - \log_{10}(x - 1) = 1$