Problem 14 - Algebra A Combined Function

Question

Solve each equation. Give an exact solution and a four-decimal-place approximation. $$ 5^{2 x-6}=12 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{2}\left(\frac{\ln(12)}{\ln(5)} + 6\right) \), approximately \( x \approx 4.2718 \).

Step 1: Isolate the Exponential Expression

First, recognize that the equation is already isolated with the exponential expression on one side: \[ 5^{2x-6} = 12 \] No further manipulation is needed to isolate the exponential.

Step 2: Apply the Logarithm Function

Apply the natural logarithm (ln) to both sides of the equation to solve for the exponent:\[ \ln(5^{2x-6}) = \ln(12) \]Use the property of logarithms that allows the exponent to be brought in front:\[ (2x-6) \cdot \ln(5) = \ln(12) \]

Step 3: Solve for the Expression with x

Divide both sides of the equation by \(\ln(5)\) to isolate the expression containing \(x\):\[ 2x - 6 = \frac{\ln(12)}{\ln(5)} \]

Step 4: Isolate x

Add 6 to both sides:\[ 2x = \frac{\ln(12)}{\ln(5)} + 6 \]Then divide by 2:\[ x = \frac{1}{2}\left(\frac{\ln(12)}{\ln(5)} + 6\right) \]

Step 5: Calculate the Exact Solution

At this point, calculate the exact value:\[ x = \frac{1}{2}\left(\frac{\ln(12)}{\ln(5)} + 6\right) \]Do not simplify further as we want an expression that uses common logarithms.

Step 6: Calculate the Approximate Value

Use a calculator to find the decimal value:\( \ln(12) \approx 2.4849 \) and \( \ln(5) \approx 1.6094 \). Substitute these into the equation:\[ x \approx \frac{1}{2}\left(\frac{2.4849}{1.6094} + 6\right) \] Calculate the result to four decimal places:\[ x \approx 4.2718 \]

Key Concepts

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are a type of logarithm with base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. They are particularly useful in calculus and advanced mathematics because they simplify many equations, especially those involving growth processes like compound interest and radioactive decay.
In the given problem, applying natural logarithms to both sides of the equation \( 5^{2x-6} = 12 \), allows us to bring the exponent \( 2x-6 \) down as a coefficient. This simplification is crucial because it turns the exponential equation into a more manageable linear form:

• \( \ln(5^{2x-6}) = \ln(12) \)
• Using logarithmic identity: \( (2x-6) \cdot \ln(5) = \ln(12) \)

Understanding this process is essential for solving exponential equations efficiently.

Exact Solutions

An exact solution in mathematics denotes a solution that is expressed in terms of known numbers or mathematical constants, without approximation. For the equation \( 5^{2x-6} = 12 \), the exact solution does not include any decimal approximations. Instead, it's given in terms of logarithmic expressions and constants.
The critical steps in obtaining an exact solution are as follows:

• Building the equation: \( (2x-6) \cdot \ln(5) = \ln(12) \)
• Rearranging for \( x \): \( x = \frac{1}{2}\left(\frac{\ln(12)}{\ln(5)} + 6\right) \)

This form retains all exactness characteristics because it leverages properties of logarithms without approximating the numerical values.

Decimal Approximation

Decimal approximation is the method of expressing a number as a finite decimal, often rounded to a specified number of decimal places. This is especially useful when exact values aren't practical for interpretation or further calculations. In the given problem, once we have the exact expression for \( x \), we must calculate its decimal approximation to four decimal places.
To achieve this:

• We first find \( \ln(12) \approx 2.4849 \) and \( \ln(5) \approx 1.6094 \)
• Substitute these values into the expression for \( x \)
• Perform the calculations: \( x \approx \frac{1}{2}\left(\frac{2.4849}{1.6094} + 6\right) \)
• Resulting in a value: \( x \approx 4.2718 \)

This approximation allows you to interpret the solution meaningfully in a more practical form.

Exponential Expression

Exponential expressions are mathematical expressions that involve a constant base raised to a variable exponent. They are crucial in modeling situations with proportional growth or decay, such as population growth or radioactive decay. In the problem \( 5^{2x-6} = 12 \), the term \( 5^{2x-6} \) is an example of an exponential expression.
Here are some important features of exponential expressions:

• The base \( 5 \) is a constant
• The exponent \( 2x-6 \) makes the expression depend on the variable \( x \)
• Solve such expressions involves isolating the exponential part before applying logarithmic functions to linearize it

Understanding these attributes assists in translating the equation into one that's feasible to solve using techniques like logarithms.