Problem 17 - Precalculus

Question

Solve the equation analytically. $$ \log _{169}(3 x+7)-\log _{169}(5 x-9)=\frac{1}{2} $$

Step-by-Step Solution

Answer

The solution is \(x = 2\).

Step 1: Apply Logarithmic Property

First, we use the property of logarithms that states \( ext{log}_b(a) - ext{log}_b(c) = ext{log}_b\left(\frac{a}{c}\right)\). This allows us to combine the logarithms into one. So, \(\log_{169}(3x+7) - \log_{169}(5x-9)\) becomes \(\log_{169}\left(\frac{3x+7}{5x-9}\right)\).

Step 2: Exponential Form Conversion

Next, we convert the logarithmic equation \(\log_{169}\left(\frac{3x+7}{5x-9}\right) = \frac{1}{2}\) into its exponential form. Recall that \(\log_b(a) = c\) implies \(a = b^c\). Thus, \(\frac{3x+7}{5x-9} = 169^{\frac{1}{2}}\).

Step 3: Simplify the Exponential Expression

Calculate \(169^{\frac{1}{2}}\), which is the square root of 169. Therefore, \(169^{\frac{1}{2}} = 13\). Now, our equation simplifies to \(\frac{3x+7}{5x-9} = 13\).

Step 4: Solve the Linear Equation

Now, cross-multiply to eliminate the fraction: \(3x + 7 = 13(5x - 9)\). Distribute the 13 on the right side: \(3x + 7 = 65x - 117\).

Step 5: Isolate \(x\)

Subtract \(3x\) from both sides to get \(7 = 62x - 117\). Then add 117 to both sides to get \(124 = 62x\). Finally, divide by 62 to solve for \(x\): \(x = \frac{124}{62} = 2\).

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools for simplifying and solving logarithmic equations. One useful property is the difference of logs:

• \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \)

By applying this property, you can compress two separate logarithms into a single log expression. This is particularly useful for cases where the base of the logarithms is the same, as seen in our original exercise.
In the problem

• \( \log_{169}(3x+7) - \log_{169}(5x-9) = \frac{1}{2} \)

we can simplify it using the property to

• \( \log_{169}\left(\frac{3x+7}{5x-9}\right) = \frac{1}{2} \)

This makes the equation much easier to handle, as it consolidates multiple logarithmic expressions into a single one, allowing further manipulation.

Exponential Form

Converting logarithmic equations to exponential form is an effective way to solve them. The basic idea here is to use the relationship between logs and exponents. For a given logarithmic equation:

• \( \log_b(a) = c \)

you can translate it to its exponential equivalent:

• \( a = b^c \)

In our situation, this means translating:

• \( \log_{169}\left(\frac{3x+7}{5x-9}\right) = \frac{1}{2} \)

into:

• \( \frac{3x+7}{5x-9} = 169^{\frac{1}{2}} \)

Through this conversion, you turn the logarithm into a more straightforward expression involving powers, which simplifies subsequent calculations.

Solving Linear Equations

Once you have an equation set up without logs, the next step is to simplify it by solving the linear equation formed. This involves:

• Expanding the expressions.
• Simplifying terms.
• Finding the variable \( x \).

Looking at the equation

• \( \frac{3x+7}{5x-9} = 13 \)

you proceed by cross-multiplying to eliminate the fraction and simplify the equation.
The equation's transformation to

• \( 3x + 7 = 13(5x - 9) \)

yields a simple linear equation:

• \( 3x + 7 = 65x - 117 \)

From here, isolate \( x \) through basic algebraic manipulations to arrive at the solution.

Cross Multiplication

Cross multiplication is a technique used to clear fractions from equations, particularly useful when dealing with proportions. By using this method, you can transform a fractional equation into a simpler, non-fractional one. The method involves:

• Multiplying across the equals sign diagonally.
• Clearing both sides of the fraction simultaneously.

Taking the equation

• \( \frac{3x+7}{5x-9} = 13 \)

we cross multiply to get:

• \( 3x + 7 = 13 \times (5x - 9) \)

This results in:

• \( 3x + 7 = 65x - 117 \)

Once multiplied, it eliminates the fraction, making it easier to solve for \( x \) using standard algebraic techniques. Cross multiplication is thus indispensable in the process of solving proportion-related problems efficiently.