Solving Exponential Equations -15 * 2^(-0.5x) = -90

Have you ever encountered an equation that looks intimidating at first glance? Exponential equations, with their variables lurking in the exponents, can seem like mathematical puzzles. But fear not! With a few key steps and a trusty calculator, even the most complex-looking exponential equations can be tamed. In this article, we'll dissect the equation $-15 imes 2^{-0.5 x}=-90$, unraveling its secrets and revealing the solution. So, grab your calculator, and let's embark on this mathematical adventure!

Unveiling the Exponential Equation

At its heart, an exponential equation is simply an equation where the variable appears in the exponent. These equations often model real-world phenomena like population growth, radioactive decay, and compound interest, making them essential tools in various fields. The equation $-15 imes 2^{-0.5 x}=-90$ is a classic example, where 'x' resides in the exponent of the base 2. Our mission is to isolate 'x' and determine its value.

Isolating the Exponential Term: The First Step

The golden rule of equation solving is to isolate the variable we're after. In this case, our target is the exponential term, $2^{-0.5x}$. To achieve this, we need to rid of the coefficient -15 that's clinging to it. A simple division operation will do the trick. Dividing both sides of the equation by -15, we get:

$$2^{-0.5 x} = 6$$

Now, the exponential term stands alone, ready for the next stage of our solution journey.

The Logarithmic Bridge: Crossing into the Exponent's Domain

With the exponential term isolated, we face a new challenge: how do we bring the exponent, -0.5x, down from its lofty perch? This is where logarithms, the inverse operations of exponentiation, enter the scene. Logarithms act as a bridge, allowing us to cross into the exponent's domain. The key is to apply the logarithm to both sides of the equation. We can use any base for the logarithm, but the common logarithm (base 10) or the natural logarithm (base e) are often the most convenient, as they are readily available on calculators.

Let's wield the natural logarithm (ln) in our quest. Applying the natural logarithm to both sides of the equation, we get:

$$ln(2^{-0.5 x}) = ln(6)$$

Unleashing the Power Rule: Taming the Exponent

The magic of logarithms lies in their ability to transform exponents into coefficients. The power rule of logarithms states that $ln(a^b) = b imes ln(a)$. Applying this rule to our equation, we can bring the exponent -0.5x down as a coefficient:

$$-0.5 x imes ln(2) = ln(6)$$

Now, the variable 'x' is no longer trapped in the exponent; it's free to be isolated!

The Final Stretch: Isolating 'x' and Finding the Solution

With 'x' liberated from the exponent, the final steps are straightforward. We have the equation:

$$-0.5 x imes ln(2) = ln(6)$$

To isolate 'x', we need to divide both sides by $-0.5 imes ln(2)$: $rac{-0.5 x imes ln(2)}{-0.5 imes ln(2)} = rac{ln(6)}{-0.5 imes ln(2)}$ This simplifies to:

x = rac{ln(6)}{-0.5 imes ln(2)}

Now, it's calculator time! Evaluate the right-hand side of the equation, and we get:

$$x imes -5.170$$

Rounding to the nearest thousand, we have our solution: $x ≈ -5.170$

Reflecting on the Solution

The solution, approximately -5.170, is the value of 'x' that satisfies the original equation. Plugging this value back into the equation $-15 imes 2^{-0.5 x}=-90$ should yield a true statement (within the margin of rounding errors). It's always a good practice to verify your solution to ensure accuracy.

Key Concepts in Solving Exponential Equations

To solidify your understanding, let's recap the key concepts we've employed in solving this exponential equation:

1. Isolating the Exponential Term: This is the crucial first step, setting the stage for applying logarithms.
2. The Logarithmic Bridge: Logarithms are the key to unlocking the exponent, allowing us to bring it down as a coefficient.
3. The Power Rule: This rule, $ln(a^b) = b imes ln(a)$, is the magic wand that transforms exponents into coefficients.
4. Isolating the Variable: Once the exponent is tamed, isolating the variable 'x' becomes a matter of basic algebraic manipulation.

With these concepts in your arsenal, you'll be well-equipped to tackle a wide range of exponential equations.

Different Approaches to the Same Destination

While we've used the natural logarithm (ln) to solve this equation, it's worth noting that we could have used the common logarithm (log base 10) or any other base. The choice of base doesn't affect the final solution, but some bases might simplify the calculations in certain situations. For instance, if the base of the exponential term was 10, using the common logarithm would be a natural choice. To illustrate this, let's solve the same equation using the common logarithm:

Original equation: $-15 imes 2^{-0.5 x}=-90$

Isolating the exponential term: $2^{-0.5 x} = 6$

Applying the common logarithm: $log(2^{-0.5 x}) = log(6)$

Using the power rule: $-0.5 x imes log(2) = log(6)$

Isolating 'x': $x = rac{log(6)}{-0.5 imes log(2)}$

Evaluating with a calculator, we arrive at the same solution: $x ≈ -5.170$

This demonstrates the flexibility of logarithms in solving exponential equations. The key is to choose the base that feels most comfortable and efficient for the given problem.

Real-World Applications of Exponential Equations

Exponential equations aren't just abstract mathematical constructs; they have profound implications in various real-world scenarios. Let's explore a few examples:

• Population Growth: Exponential equations model how populations grow over time, assuming a constant growth rate. Understanding population dynamics is crucial for resource management, urban planning, and environmental conservation.
• Radioactive Decay: The decay of radioactive isotopes follows an exponential pattern. This principle is used in carbon dating, a technique for determining the age of ancient artifacts and fossils.
• Compound Interest: The growth of investments with compound interest is governed by exponential equations. Understanding compound interest is essential for financial planning and wealth accumulation.
• Spread of Diseases: The initial spread of infectious diseases can often be modeled using exponential functions. This knowledge helps epidemiologists predict outbreaks and implement control measures.
• Cooling and Heating: The rate at which an object cools or heats up is often proportional to the temperature difference between the object and its surroundings, leading to exponential models.

These examples highlight the power and versatility of exponential equations in describing and predicting real-world phenomena. Mastering the techniques for solving these equations opens doors to understanding and analyzing a wide range of processes.

Conclusion: Empowered to Solve

Exponential equations, while initially daunting, are ultimately conquerable with the right tools and techniques. By isolating the exponential term, applying logarithms, wielding the power rule, and isolating the variable, we can unlock the solutions hidden within these equations. The equation $-15 imes 2^{-0.5 x}=-90$ served as a valuable example, illustrating the step-by-step process of solving exponential equations. Remember, practice makes perfect! The more you engage with these equations, the more confident and proficient you'll become in solving them. So, embrace the challenge, grab your calculator, and embark on your own exponential equation-solving adventures!

What is the solution of the equation? Round your answer, if necessary, to the nearest thousand.

To find the solution to the equation $-15 imes 2^{-0.5 x}=-90$, we followed a step-by-step process that involved isolating the exponential term, applying logarithms, using the power rule, and finally, isolating the variable 'x'.

First, we isolate the exponential term by dividing both sides of the equation by -15:

$$2^{-0.5 x} = 6$$

Next, we apply the natural logarithm (ln) to both sides of the equation:

$$ln(2^{-0.5 x}) = ln(6)$$

Using the power rule of logarithms, we bring the exponent down as a coefficient:

$$-0.5 x imes ln(2) = ln(6)$$

Now, we isolate 'x' by dividing both sides by $-0.5 imes ln(2)$:

x = rac{ln(6)}{-0.5 imes ln(2)}

Evaluating this expression with a calculator, we get:

$$x imes -5.170$$

Rounding to the nearest thousand, the solution to the equation is approximately:

$$x ≈ -5.170$$

Therefore, the solution of the equation $-15 imes 2^{-0.5 x}=-90$, rounded to the nearest thousand, is approximately -5.170.