Problem 84 - Calculus for Biology and Medicine

Question

Solve for \(x\). (a) \(\log _{3}(2 x-1)=2\) (b) \(\ln (2-3 x)=0\) (c) \(\log (x)-\log (x+1)=\log \left(\frac{2}{3}\right)\)

Step-by-Step Solution

Answer

(a) x = 5, (b) x = \(\frac{1}{3}\), (c) x = 2.

Step 1: Identify the Equation Type

Recognize that the problem involves logarithmic equations, making use of properties of logarithms to solve for the variable.

Step 2: Solve Part (a) - Convert Logarithm to Exponential Form

The equation is \( \log_{3}(2x-1) = 2 \). Rewriting in exponential form gives: \( 2x-1 = 3^2 \).

Step 3: Solve Part (a) - Solve for x

Calculate \( 3^2 = 9 \). Thus, the equation \(2x-1 = 9\) becomes \(2x = 10\). Solving for \(x\) gives \(x = 5\).

Step 4: Solve Part (b) - Use Exponential Form of Natural Logarithm

The equation is \( \ln(2-3x) = 0 \). Convert to exponential form: \( 2-3x = e^0 \).

Step 5: Solve Part (b) - Solve for x

Since \( e^0 = 1 \), we have \( 2-3x = 1 \). Solve this to find \(x = \frac{1}{3} \).

Step 6: Solve Part (c) - Use Logarithm Properties

The equation is \( \log(x) - \log(x+1) = \log\left(\frac{2}{3}\right) \). Use the quotient rule: \( \log\left(\frac{x}{x+1}\right) = \log\left(\frac{2}{3}\right) \).

Step 7: Solve Part (c) - Remove the Logarithms and Solve for x

By setting the arguments equal, \( \frac{x}{x+1} = \frac{2}{3} \). Cross-multiply to get \( 3x = 2(x+1) \). Simplify to \(3x = 2x + 2\) and solve for \(x\) to find \(x = 2\).

Key Concepts

Properties of Logarithms

Logarithms have their unique set of properties that can help simplify and solve equations. When you encounter logarithmic equations, these properties become especially handy. Here are the key properties used in solving logarithmic equations:

• Product Rule: This rule states that the logarithm of a product is equal to the sum of the logarithms of the factors. In formula terms, \( \log_b(MN) = \log_b(M) + \log_b(N) \).
• Quotient Rule: Similarly, the logarithm of a quotient is equal to the difference of the logarithms. It is expressed as \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \).
• Power Rule: This property tells us that the logarithm of a power is the exponent times the logarithm of the base. Represented as \( \log_b(M^p) = p\log_b(M) \).

These rules allow you to combine or separate logarithmic expressions into simpler forms, making it easier to solve for the variable. In Exercise Part (c), the quotient rule was utilized to simplify the logarithmic equation \( \log(x) - \log(x+1) = \log\left(\frac{2}{3}\right) \), by rewriting it as a single logarithm.

Exponential Form

Converting a logarithmic equation into an exponential form is a crucial step in solving such equations. This transformation allows you to work directly with powers and exponential expressions, which can often be more straightforward. The essential idea is that if you have a log equation like \( \log_b(A) = C \), you can rewrite it as its equivalent exponential form: \( A = b^C \). This is because a logarithm tells you what power you would raise the base \( b \) to get the number \( A \). In Exercise Part (a), the equation \( \log_{3}(2x-1) = 2 \) is transformed into \( 2x-1 = 3^2 \) using the exponential form. On calculating \( 3^2 \), you then have a simple linear equation, \( 2x-1 = 9 \), which can be solved easily for \( x \). Similarly, in Part (b), \( \ln(2-3x) = 0 \) becomes \( 2-3x = e^0 \) or simply \( 2-3x = 1 \), enabling straightforward calculation of the value of \( x \).

Solving Equations

Once you have simplified or transformed the equations using the properties of logarithms or exponential form, the remaining task is to solve them. Solving equations involves basic algebraic techniques that allow you to find the unknown variable. For linear equations, such as those obtained in parts (a) and (b) of the exercise, the process generally involves isolating the variable on one side of the equation. For example:

• For \( 2x-1 = 9 \), you add 1 to both sides to get \( 2x = 10 \) and then divide by 2 to find \( x = 5 \).
• For \( 2-3x = 1 \), subtract 2 from both sides to get \( -3x = -1 \) and divide by -3 to get \( x = \frac{1}{3} \).

In Part (c) of the exercise, you end up with a rational equation \( \frac{x}{x+1} = \frac{2}{3} \). To solve it, you cross-multiply to eliminate the fractions: \( 3x = 2(x + 1) \). Expand and simplify to find the solution for \( x \). By consistently applying these algebraic methods, solving logarithmic equations becomes a systematic process.