91Ó°ÊÓ

Exponential equations involve expressions where the variable is an exponent. In the context of our exercise, we first needed to transform the given logarithmic equation into an exponential form. The equation provided was \[\log_{3}(4c+5) = 3\]. To rewrite it exponentially, we applied the basic concept that if \[\log_{b}(x) = y\], then \[b^y = x\]. Such transformations are crucial because they enable us to work with simpler arithmetic and algebraic steps.
In our case:

• Base \(b\) is 3
• The argument \(x\) is \((4c+5)\)
• And the result \(y\) is 3.

Thus, when rewritten, we get \[3^3 = 4c + 5\], which simplifies to \[27 = 4c + 5\]. Working with exponential equations often involves simplification, allowing us to isolate terms involving the unknown variable.

Logarithms are incredibly useful in solving equations where the unknown variable is an exponent. Understanding the properties of logarithms helps in converting complex expressions into more manageable forms.
For this exercise, the property of interest is the conversion from logarithmic to exponential form, which is central to solving the equation \[\log_{3}(4c+5)=3\]. This transformation relies on the fundamental property: if \[\log_{b}(x) = y\], then it implies \[b^y = x\].
Why is this important? It turns challenging logarithmic equations into solvable polynomial equations. This step is essential as it translates a logarithm-based problem into a form where algebraic manipulation can be employed easily, making it straightforward to isolate variables and solve for unknowns.Furthermore, understanding properties, such as the product, quotient, and power rules of logarithms, could be useful in other contexts but are not directly applied in this particular problem.

Algebraic manipulation involves rewriting equations and expressions to simplify or solve them. Once our equation is in the exponential form, the main task is to isolate the variable. This means ensuring the variable is by itself on one side of the equation.
For the given problem, once we had \[27 = 4c+5\] as our working equation, it became important to manipulate it to solve for \(c\).

• Subtracting 5 from both sides to get rid of the constant term: \[27 - 5 = 4c\]
• This simplifies to \[22 = 4c\]
• Then, dividing both sides by 4 enables us to isolate \(c\): \[c = \frac{22}{4}\]

Finally, simplifying the fraction gives us \[c = \frac{11}{2}\]. Such methods—step-by-step subtraction, division, and simplification—are foundational tools in algebra for solving equations and ensuring expressions are in their simplest form. Mastering these techniques allows for efficient and accurate solutions to many types of algebraic problems.