Problem 34 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(8\left(4^{6-2 x}\right)+13=41\)

Step-by-Step Solution

Answer

Calculating this equation yields an approximate result of \(x \approx 2.183\) when rounded to 3 decimal places.

Step 1: Isolate the Exponential Function

Subtract 13 from both sides of the equation to isolate the exponential term on one side. This gives: \(8\left(4^{6-2 x}\right) = 41 - 13 = 28\) which simplifies to \(4^{6-2 x} = \frac{28}{8} = 3.5\)

Step 2: Make Use of Logarithm Properties

Logarithmic properties can be used to isolate the x by applying the natural log (ln) to both sides of the equation. So, \(\ln(4^{6-2 x}) = \ln(3.5)\). This can further be simplified due to the properties of logs into \((6-2x)\ln(4) = \ln(3.5)\). This essentially allows us to remove x from the exponent and work with it in the base of the equation.

Step 3: Solve for x

Isolate x by first subtracting \(6\ln(4)\) from both sides, then divide both sides by \(-2\ln(4)\) to get: \(x = \frac{\ln(3.5) - 6\ln(4)}{-2\ln(4)}\)

Key Concepts

Exponential Function Isolation

When solving exponential equations like the one given, the first step is to isolate the exponential function. Isolating the exponential component from other terms is crucial, as it makes the application of logarithms feasible, which is a key step in solving such equations. The process of isolation often involves basic algebraic operations, such as addition, subtraction, multiplication, and division. In the original exercise, we subtract 13 from both sides to isolate the term involving the exponential function, transforming the equation into a simpler form where the exponential is clearly presented and ready for further operations.

Logarithmic Properties

Once the exponential function is isolated, logarithmic properties become essential in solving the equation. Logarithms can transform multiplicative processes into additive ones, which is incredibly useful when dealing with exponents. In the context of the given problem, applying the natural logarithm to both sides of the equation (a process known as 'logarithmizing') allows us to use the property that \(\ln(a^b) = b \cdot \ln(a)\). This key property lets us bring down the exponent and set the stage for solving for the unknown variable. Since logarithms are inverses of exponentials, they are the most effective tool for dealing with exponential equations.

Algebraic Manipulation

To find the value of 'x', algebraic manipulation comes into play after logarithmic properties have been applied. It involves reordering and simplifying the equation to isolate 'x' and make it the subject of the formula. These manipulations include distributing the logarithm across the terms in the exponent, collecting like terms, and then dividing both sides of the equation by a coefficient to solve for 'x'. In the exercise, subtracting and dividing by logarithmic terms illustrates algebraic manipulation. Such steps are indispensable to transition from a complex equation to a more straightforward one that reveals the value of the unknown variable.

Natural Logarithm Application

The natural logarithm, symbolized as 'ln', is specifically used to solve equations involving the natural exponential function 'e'. However, it can also be applied to other bases, as seen in the given equation. Its application is particularly important because it simplifies expressions involving 'e' and allows for the solution of exponents in the equation. By taking the natural log of both sides of the equation, we leverage the unique capability of ln to deal with exponents effectively, bringing the variable out of the exponent and enabling us to solve for it algebraically. In the solution process, once we have applied the natural logarithm and used its properties, we can reach the stage where the variable we need to find is no longer in the exponent.
This effectively demystifies the process of solving exponential equations and brings us closer to the final solution.