Problem 25 - Precalculus : Building Concepts and Connections

Question

Solve the exponential equation. Round to three decimal places, when needed. $$2\left(0.8^{x}\right)-3=8$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{ln(5.5)}{ln(0.8)}\), which is approximately 3.274 when rounded to three decimal places.

Step 1: Isolate the exponential part

We should start by isolating \(0.8^{x}\). Doing that, we get \(0.8^{x}\) by adding 3 to both sides and then dividing by 2: \[2\left(0.8^{x}\right)=8+3\] which simplifies to \(2\left(0.8^{x}\right)=11\] Then divide both sides by 2 to get: \[0.8^{x} = \frac{11}{2} = 5.5\]

Step 2: Apply the logarithm

To get rid of the exponent, we apply the natural logarithm (ln) to both sides of the equation. We can use the property of logarithms, \(ln(a^b) = b* ln(a)\). Thus, we have: \[ln(0.8^{x}) = ln(5.5)\] which simplifies to \(x* ln(0.8) = ln(5.5)\]

Step 3: Solve for x

Now, we can solve for \(x\). To do this, we divide both sides by \(ln(0.8)\): \[x = \frac{ln(5.5)}{ln(0.8)}\] Calculating the right side of the equation gives the numerical value for \(x\). Depending upon the calculator, this may differ slightly.

Step 4: Round the answer

As per the question's requirement, the answer should be rounded to three decimal places. Remember that rounding should be done at the very end of the calculation to avoid rounding errors.

Key Concepts

Logarithms

Logarithms are mathematical operations that help us solve equations where the variable is an exponent. For example, in the equation \(0.8^x = 5.5\), the exponent \(x\) is the variable we aim to solve for. Using logarithms allows us to "bring down" the exponent in front of the logarithm, making it easier to solve for the variable.

Logarithms have a special property: \(\log_b(a^c) = c \cdot \log_b(a)\). This property is essential in solving exponential equations. In our exercise, we apply the natural logarithm to both sides of the equation and use this property to isolate the variable \(x\).

• Logarithms transform multiplicative processes into additive ones.
• They are used to "untangle" power-based functions.
• The logarithm of a number to a base tells you the exponent needed to obtain that number.

Solving equations using logarithms is common in fields such as physics, computer science, and finance.

Natural Logarithm

The natural logarithm, denoted as \(ln\), is a specific type of logarithm with a base \(e\), where \(e\) is approximately 2.71828. It is particularly useful in calculus and many scientific formulas because \(e\) appears naturally in various mathematical contexts.

When dealing with exponential equations, using the natural logarithm can simplify calculations. In the context of our problem, we used \(ln(0.8^x) = ln(5.5)\). By applying \(ln\) to both sides, we convert the exponential equation into a form where the variable \(x\) can be isolated easily.

• The natural logarithm is used extensively in growth and decay problems.
• Systems involving continuous rates of change often use \(ln\).
• Manipulating exponential equations becomes more manageable with natural logarithms.

Because of its properties, the natural logarithm is a powerful tool in both theoretical and applied mathematics.

Isolating the Variable

Isolating the variable in an equation means rewriting the equation so that the variable appears by itself on one side of the equation. This process is crucial when solving exponential equations, as it allows you to determine the exact value of the variable.

In our given exercise, we initially isolated the term \(0.8^x\) by performing operations to remove other terms from its side of the equation. For instance, we added 3 to both sides and then divided by 2 to separate the power term \(0.8^x\) entirely.

• Start by performing arithmetical operations to leave the variable term alone.
• Use properties of logarithms to further simplify and isolate the variable.
• Adjust both sides equally to maintain the equation's balance.

This step-by-step isolation helps simplify the equation and is vital for accurately finding the solution, especially when dealing with exponential terms.