Question

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

Short answer

After calculating the right side of the equation, the approximate value of 'x' is found to be 1.466 to three decimal places.

Step-by-step solution

Simplify Equation

Start by subtracting 13 from both sides to isolate the term with the exponential expression. This would produce a new equation: \(8\left(4^{6-2 x}\right) = 41 - 13 = 28\)

Further Isolate Exponential Term

Divide both sides of the equation by 8 to further isolate the exponential term. This yields \(4^{6-2x} = \frac{28}{8} = 3.5\)

Use Logarithm to Eliminate Exponent

Use natural logarithm on both sides to let you eliminate the exponent. This results in: \(ln\left(4^{6-2x}\right) = ln(3.5)\), which simplifies to \((6 - 2x)ln(4) = ln(3.5)\)

Solve for x

Lastly, solve the resulting equation for x. This results in: \(x = \frac{6 - ln(3.5)}{2ln(4)}\)

Key concepts

Logarithms

Logarithms, often abbreviated as logs, are the inverse operations of exponentials. If you have an exponential equation like \(4^{b} = a\), it can be rewritten as a logarithmic equation, which is \(b = \log_{4}(a)\). Essentially, logarithms help in answering the question "to what power must the base be raised, to yield the given number?"
In problems involving exponential equations, logs are especially useful for isolating variables that appear as exponents. The natural logarithm, denoted as \(\ln\), uses the constant \(e\) as its base, but functions similarly to other logarithmic forms. When taking the log of both sides of an equation, you essentially use this property to "bring down" exponents, making it easier to solve for the variable inside the exponent.
In our original problem, by using the natural logarithm, \(\ln\), we were able to convert the equation \(4^{6-2x} = 3.5\) into a form that could be solved for \(x\) as \((6 - 2x)\ln(4) = \ln(3.5)\). Remember, the logarithmic identity \(\ln(a^b) = b\ln(a)\) allows us to move the exponent out in front of the logarithm, turning our problem into an algebraic one.

Algebraic Manipulation

Algebraic manipulation refers to the variety of operations that you can perform to simplify or solve equations. It involves addition, subtraction, multiplication, and division, among other operations, to rearrange expressions and isolate the desired variable.
In the step-by-step solution of our exercise, we started by isolating the term with the exponential expression. Initially, 13 was subtracted from both sides in the equation \(8(4^{6-2x}) + 13 = 41\) to give us \(8(4^{6-2x}) = 28\). It's important to perform the same operation on both sides to maintain equality.
Further manipulation involved dividing both sides by 8, resulting in \(4^{6-2x} = 3.5\). This step was crucial for isolating the exponential term and preparing for the application of logarithms. By breaking equations down into simpler parts, algebraic manipulation makes it easier to solve for the unknowns.

Solving Equations

Solving equations involves finding the value of unknown variables that make an equation true. This process often combines several mathematical techniques to isolate the variable and solve for its value.
In the exponential equation given, solving for \(x\) involved a sequence of strategic steps. After simplifying the equation to \(4^{6-2x} = 3.5\) through algebraic manipulation, logarithms were employed to handle the exponent. Applying the natural logarithm to both sides, we obtained \((6 - 2x)\ln(4) = \ln(3.5)\), which allowed for further simplification.
The final step in finding \(x\) focused on isolating \(x\) itself: rearranging the equation \((6 - 2x)\ln(4) = \ln(3.5)\) to \(x = \frac{6 - \ln(3.5)}{2\ln(4)}\) provided the solution. Solving equations requires a methodical approach, patiently working through each step until the variable is isolated and known.