Problem 33 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each exponential equation by taking the logarithm on both sides. Express the solution set in terms of logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$30-(1.4)^{x}=0$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = \ln(30)/\ln(1.4)\) which approximates to \(x \approx 16.91\) when rounded to two decimal places.

Step 1: Format the Equation

First, we rearrange the given equation \(30-(1.4)^{x}=0\) in order to isolate the exponential term. This gives us \( (1.4)^{x} = 30 \).

Step 2: Taking logarithm on both sides

Now, we will take the natural logarithm (or any base of your choice, here we will use natural log for easier computation later) on both sides of the equation. This gives \(\ln((1.4)^{x}) = \ln(30)\).

Step 3: Simplifying the Equation

Next, we make use of the property of logarithms that allows us to bring the exponent out in front. So our equation becomes \(x \* \ln(1.4) = \ln(30)\).

Step 4: Solving for the variable

Then we solve for \(x\) to express the solution in terms of logarithms by dividing both sides of our equation by \(\ln(1.4)\). This gives \(x = \ln(30)/\ln(1.4)\).

Step 5: Decimal Approximation

Finally, we use a calculator to find a decimal approximation for the solution, rounding to two decimal places. The calculator provides \(x \approx 16.91\).

Key Concepts

Logarithms

Logarithms are mathematical tools used to solve exponential equations like the one given in the exercise. They allow us to reverse the operation of exponentiation, making it easier to find unknown variables that appear as exponents. In simple terms, a logarithm answers the question: "To what power do we raise a specific base to get a certain number?" For example, if you have \(b^y = x\), the logarithm of \(x\) with base \(b\) is \(y\), expressed as \( ext{log}_b(x) = y\). To solve exponential equations, you can take the logarithm of both sides of the equation.
This effectively "frees" the variable from the exponent slot and brings it down to the level of multiplication, which is easier to handle mathematically.
In our specific example, the equation \(30 - (1.4)^x = 0\) was transformed to \( (1.4)^x = 30\), and logarithms were used to express \(x\) as \(x = rac{ ext{ln}(30)}{ ext{ln}(1.4)}\). This process highlights the usefulness of logarithms in solving equations that may at first glance seem difficult to decipher.

Decimal Approximation

Decimal approximation is a method used to express answers in a more understandable form, especially in mathematics where more precise values are not always necessary or available. In the original exercise, after deriving the equation \(x = rac{ ext{ln}(30)}{ ext{ln}(1.4)}\), the next step was to find the decimal approximation for \(x\).
Here, we used a calculator to compute the natural logarithms and their division, providing a numerical result of \(x \approx 16.91\).
Decimal approximations are particularly useful in real-life applications where exact values can be unwieldy or not practical to use. In this case, rounding the result to two decimal places gives a balance between precision and simplicity. Tools like calculators are indispensable for these computations, ensuring results are accurate and helping with verification of mathematical work.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a specific type of logarithm with the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. Natural logarithms are widely used due to their connection to continuous growth and natural processes, making them pivotal in fields such as calculus and financial modeling.
In the process of solving exponential equations, using the natural logarithm can simplify calculations, especially when dealing with the derivative or integral of functions. In our example exercise, the natural logarithm was chosen to simplify obtaining the solution for \(x\).

• We took the natural logarithm of both sides of the equation: \(\ln((1.4)^x) = \ln(30)\).
• This makes it straightforward to use the properties of logarithms to manage the expression.
• It converted the exponential equation into a linear format: \(x \cdot \ln(1.4) = \ln(30)\), from which \(x\) was isolated.

Understanding the natural logarithm helps students tackle problems involving exponential functions efficiently and is often essential in advanced mathematics.