Problem 66 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the exponential equation algebraically. Round your result to three decimal places. Use a graphing utility to verify your answer. $$6\left(8^{-2-x}\right)+15=2601$$

Step-by-Step Solution

Answer

Solving the given exponential equation and rounding off to three decimal places gives the solution \(x \approx -7.52\) after substituting it back into the original equation for testing validity.

Step 1: Isolate the term with the variable in exponent

First subtract 15 from both sides of the equation to isolate the term containing exponent.The equation then becomes \(6\left(8^{-2-x}\right)=2601-15, which simplifies to 6\left(8^{-2-x}\right)=2586.

Step 2: Remove the coefficient in front of the exponential term

Now divide both sides of the equation by 6. Therefore, our equation becomes\(8^{-2-x}=2586/6, which simplifies to 8^{-2-x} =431.

Step 3: Apply Logarithm to both sides

To remove the exponential, we can take the logarithm on both sides. We take the logarithm base 8 on both sides of the equation. This gives us \(-2-x = \log_8{431}

Step 4: Solve for x

Now, the variable \(x\) can be easily solved by rearranging the equation. Therefore, \(x = -2 - \log_8{431}\)

Step 5: Round off the number

Upon performing above calculations, the solution is approximately \(x = -7.52\) when rounded to three decimal places

Step 6: Verify the result using a Graphing Utility

Plot the equation \(6\left(8^{-2-x}\right)+15=2601\) and verify the value of \(x = -7.52\) by looking at the root of the equation graphically.

Key Concepts

Understanding Logarithmic Functions

To tackle equations involving exponents, one must be well-versed with logarithmic functions. A logarithm answers the question: 'To what exponent do we need to raise a base number to obtain another number?' For example, the logarithm of 100 to base 10, denoted as \( \log_{10}(100) \), is 2, because \( 10^2 = 100 \).

In the context of the given problem, the logarithmic function allows us to undo the exponential operation. This is pivotal when we have an equation such as \( 8^{-2-x} = 431 \). By applying the logarithm with a base of 8 to both sides, \( \log_{8}(8^{-2-x}) = \log_{8}(431) \), the left side simplifies to \( -2-x \) because \( \log_{b}(b^{x}) = x \) by definition. Logarithms transform the problem into a much simpler linear equation, making them an essential tool in solving exponential equations.

Isolation of Variables

Isolation of the variable is a critical step in solving algebraic equations. It involves manipulating the equation in such a way that the variable we're solving for is on one side of the equation by itself. In our exercise, the aim was to isolate \( 8^{-2-x} \).

To do this, we first move all other terms away from the term with the variable in the exponent, by subtracting 15 from both sides. Next, we divide by 6 to remove the coefficient preceding our exponential term. Each operation brings us closer to having the variable \( x \) stand alone, setting the stage for us to apply logarithmic functions and ultimately solve for \( x \) with ease.

Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. This concept is frequently illustrated in real-world phenomena such as radioactive decay and depreciation of assets. In mathematical terms, exponential decay can be represented by the function \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the quantity at time \( t \), \( N_0 \) is the initial quantity, \( e \) is the base of the natural logarithm, and \( k \) is the decay constant.

In the exercise provided, \( 8^{-2-x} \) is an exponential decay function where \( 8 \) is the base and \( -2-x \) is the exponent expressing the decay. As \( x \) increases, the value of \( 8^{-2-x} \) quickly approaches zero, illustrating the characteristic rapid decrease of exponential decay functions.