Solve each equation.
$48=6e^{4k}$

Verified step by step guidance

1. Start by isolating the exponential term. Divide both sides of the equation by 6 to get: \( e^{4k} = \frac{48}{6} \).
2. Simplify the right side of the equation: \( e^{4k} = 8 \).
3. To solve for \( k \), take the natural logarithm of both sides: \( \ln(e^{4k}) = \ln(8) \).
4. Use the property of logarithms that \( \ln(e^x) = x \) to simplify the left side: \( 4k = \ln(8) \).
5. Finally, solve for \( k \) by dividing both sides by 4: \( k = \frac{\ln(8)}{4} \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in the form of f(x) = a * e^(bx), where 'e' is the base of natural logarithms, approximately equal to 2.71828. These functions model growth or decay processes and are characterized by their rapid increase or decrease. Understanding how to manipulate and solve equations involving exponential functions is crucial for solving problems like the one presented.

Logarithms

Logarithms are the inverse operations of exponentiation, allowing us to solve for the exponent in equations of the form a^x = b. The natural logarithm, denoted as ln, specifically deals with the base 'e'. In the context of the given equation, applying logarithms helps isolate the variable 'k' by transforming the exponential equation into a linear form, making it easier to solve.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate variables or solve for unknowns. This includes operations such as addition, subtraction, multiplication, division, and applying properties of equality. In the context of the given equation, effective algebraic manipulation is necessary to isolate 'e^(4k)' and subsequently apply logarithmic properties to find the value of 'k'.