Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(2^{3-x}=565\)

Short answer

The value of x rounded to three decimal places is \(x \approx 3 - \frac{\ln(565)}{\ln(2)}\). Use a calculator for the computation and to achieve the approximately final result.

Step-by-step solution

Set Up the Equation

You start with the given equation \(2^{3-x}=565\). It represents an exponential function.

Apply the Logarithm

Next step is to apply the logarithm on both sides of the equation. For the simplicity, the natural logarithm (ln) will be used, but any other logarithmic base could be applied also. So we get \(\ln(2^{3-x}) = \ln(565)\).

Use the Properties of Logarithms

Using the property of logarithms that states \(\ln(a^b) = b\ln(a)\), we can rewrite our equation as \((3-x) \ln(2) = \ln(565)\).

Simplify the Equation

Simplify the equation by dividing by \(\ln(2)\) on both sides. That will leave the variable \(x\) alone on one side of the equation. So we get \(3-x = \frac{\ln(565)}{\ln(2)}\).

Solve for x

Next, rearrange the equation to solve for \(x\). So, \(x = 3 - \frac{\ln(565)}{\ln(2)}\).

Key concepts

Exponential Function

The concept of an exponential function is a fundamental block in algebra and occurs frequently in various mathematical contexts. It is written in the form of f(x) = a^{x}, where a is the base and x is the exponent. In the context of our problem, the function is 2^{3-x}, where 2 is the base, and 3-x is the exponent. Exponential functions are characterized by a rapid increase or decrease as the variable x changes, making them suitable to describe growth or decay processes, such as population growth or radioactive decay.

An important property of exponential functions that was put into play in the solution is that they are one-to-one functions. This property allows for their inverses to be defined, leading us to the concept of logarithms, which are vital in solving equations involving exponentials when the exponent is unknown.

Natural Logarithm

Moving onto the natural logarithm, often denoted as ln, it is simply a logarithm with a base of the mathematical constant e, approximately equal to 2.71828. The purpose of using the natural logarithm in our problem is to exploit the one-to-one nature of exponential functions to isolate the variable x.

When you apply ln to both sides of an equation with an exponent, such as 2^{3-x} = 565, you essentially 'unpack' the exponent, due to the inverse relationship between exponential functions and logarithms. This relationship is a cornerstone in solving equations of this kind, as it transforms the problem from its original exponential form into one that is solvable through basic algebraic manipulation.

Properties of Logarithms

The steps used to solve the given exponential equation also put the spotlight on the properties of logarithms. One of these properties is that logarithms convert multiplication into addition, division into subtraction, and exponentiation into multiplication. Hence, ln(a^b) = b * ln(a). This power rule allows us to bring the exponent 'down to earth', so to speak, making x accessible for further operations.

By applying this property to our equation, we were able to rewrite (2^{3-x}) as (3-x) * ln(2), thus setting the stage to solve for x through algebraic techniques. Understanding and correctly applying these properties is crucial when dealing with more complex logarithmic equations, as they are tools that simplify and streamline the solving process.

Algebraic Manipulation

Finally, algebraic manipulation is the process of rearranging and simplifying mathematical expressions and equations to solve for a variable. In the solution of our exponential equation, after applying logarithms, we're left with the expression (3-x) * ln(2) = ln(565), which required algebraic manipulation. By dividing both sides by ln(2), we isolated the term (3-x), thus making x the subject of the formula.

It's key in algebra to perform the same operation on both sides of an equation to maintain its balance. Algebraic manipulation is the essence of solving equations, as it provides the tools to move pieces around like in a puzzle until the picture – the solution – becomes clear. In this problem, the final step to find x was a simple subtraction, showcasing how algebra turns complex problems into solvable puzzles.