Problem 25 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The approximate solution to the equation is \(x = 1.681\). Remember your calculator must be set to log-base 2, or alternatively, you can use the change of base formula.

Step 1: Rearrange in Logarithmic Form

First, convert the given exponential equation \(2^{3-x} = 565\) into logarithmic form to simplify further calculations. Using the definition of logarithms, this equation can be written as: \(\log_2 565 = 3 - x\).

Step 2: Isolate the Variable

Next, you need to isolate the variable x. To achieve this, mathematically manipulate the equation so that x appears on one side by itself. Do this by rearanging to: \(x = 3 - \log_2 565\).

Step 3: Calculate the Value of x

Lastly, calculating the right-hand side of the equation will provide the value of x, using a calculator take the base 2 logarithm of 565 (remember, this means 'to what power must 2 be raised to reach 565'), subtract that number from 3, and round it to three decimal places.

Key Concepts

Logarithmic Form

When faced with an exponential equation like \(2^{3-x}=565\), transforming it into logarithmic form can simplify the problem significantly.
Logarithms help us solve for the unknown exponent by converting the expression into a much simpler form. Think of it this way: logarithms are the 'opposite' operation to exponentiation, much like subtraction is to addition. They provide a powerful tool to "bring down" an exponent when it's difficult to isolate the variable.In our case, when we rearrange the equation \(2^{3-x} = 565\) to logarithmic form, it changes to \(\log_2 565 = 3 - x\).
This transformation leverages the definition of logarithms:

• The base (in this case, 2) is used in the logarithm.
• The number 565 becomes the argument of the logarithm.
• This form directly shows the relationship needed to solve for \(x\).

Solving Equations Algebraically

After converting an exponential equation to a logarithmic form, our task shifts to solving it algebraically.
It's all about isolating the variable you are dealing with, in this case, \(x\).To isolate \(x\), we must rearrange the equation \(\log_2 565 = 3 - x\) such that \(x\) appears by itself. This involves basic algebraic operations.Here's a simple breakdown of the steps:

• Add \(x\) to both sides to avoid it being negative.
• Subtract \(\log_2 565\) from 3 to move all numeric values on one side.
• This results in: \(x = 3 - \log_2 565\).

By logically manipulating the equation like this, you make \(x\) accessible for calculation. This technique is broadly applicable in more complex equations as well.

Approximations

Once we've isolated \(x = 3 - \log_2 565\), it's time to calculate its approximate value.
Approximations are crucial, especially when exact calculations are cumbersome or impossible manually.Using a calculator to find \(\log_2 565\) is often necessary, unless you have it readily memorized.
Most calculators don't handle base 2 directly, so you may need the change of base formula:

• Convert \(\log_2 565\) using common logarithms or natural logarithms:
• \(\log_2 565 = \frac{\log_{10} 565}{\log_{10} 2}\), or \(\log_2 565 = \frac{\ln 565}{\ln 2}\)

After computing \(\log_2 565\), subtract it from 3.
Rounding the result to three decimal places gives you a precise enough approximation for most practical purposes.
This approach allows you to handle complex exponentials without straining computational resources, pivotal in real-world applications.