Solving Exponential Equations With Logarithms A Step-by-Step Guide

Hey guys! Ever stumbled upon an exponential equation that seemed like a tough nut to crack? Well, you're in the right place! In this guide, we'll break down how to solve exponential equations using logarithms, express solutions in their exact form, and even use a calculator to approximate those solutions. Let's dive into an example to make things crystal clear.

Example Equation: $2^{15x+5} = 12^{14x-4}$

Let’s tackle this equation: $2^{15x+5} = 12^{14x-4}$. Our mission is to find the value of x that satisfies this equation. Exponential equations can seem daunting, but with the power of logarithms, we can transform them into something much more manageable. The key here is to understand that logarithms allow us to bring exponents down as coefficients, making the equation linear and solvable.

Step-by-Step Solution

1. Apply Logarithms to Both Sides: To kick things off, we need to apply a logarithm to both sides of the equation. You can use any base for the logarithm, but the common logarithm (base 10) or the natural logarithm (base e) are the most convenient because they are available on most calculators. For this example, let’s use the natural logarithm (ln). Applying the natural logarithm to both sides gives us:

   $\ln(2^{15x+5}) = \ln(12^{14x-4})$

   Using the logarithm power rule, which states that $\ln(a^b) = b \ln(a)$, we can rewrite the equation as:

   $(15x+5)\ln(2) = (14x-4)\ln(12)$

2. Expand and Rearrange the Equation: Now, we need to expand both sides of the equation and rearrange terms to isolate x. This involves distributing the logarithms and then grouping the terms containing x on one side and the constants on the other side.

   Expanding both sides, we get:

   $15x \ln(2) + 5 \ln(2) = 14x \ln(12) - 4 \ln(12)$

   Next, let’s rearrange the equation to get all the x terms on one side and the constant terms on the other:

   $15x \ln(2) - 14x \ln(12) = -4 \ln(12) - 5 \ln(2)$

3. Factor Out x: To further simplify, we factor out x from the left side of the equation:

   $x(15 \ln(2) - 14 \ln(12)) = -4 \ln(12) - 5 \ln(2)$

4. Isolate x: Now, we can isolate x by dividing both sides by the expression in parentheses:

   $x = \frac{-4 \ln(12) - 5 \ln(2)}{15 \ln(2) - 14 \ln(12)}$

   This is the exact solution for x. It looks a bit complex, but it’s precise.

Expressing the Exact Solution

The exact solution we found is:

$x = \frac{-4 \ln(12) - 5 \ln(2)}{15 \ln(2) - 14 \ln(12)}$

This form is crucial because it gives us the true value of x without any rounding errors. While it might not be immediately clear what numerical value this represents, it’s the most accurate representation of the solution.

Approximating the Solution with a Calculator

To get a better sense of the value of x, we can use a calculator to approximate the solution. This involves plugging in the natural logarithms and performing the arithmetic.

1. Calculate the Logarithms: First, we need to find the values of $\ln(12)$ and $\ln(2)$ using a calculator:

   $\ln(12) \approx 2.4849$

   $\ln(2) \approx 0.6931$

2. Substitute and Compute: Now, substitute these values into the exact solution:

   $x \approx \frac{-4(2.4849) - 5(0.6931)}{15(0.6931) - 14(2.4849)}$

   $x \approx \frac{-9.9396 - 3.4655}{10.3965 - 34.7886}$

   $x \approx \frac{-13.4051}{-24.3921}$

3. Final Approximation: Divide the numerator by the denominator to get the approximate value of x:

   $x \approx 0.5496$

   So, the approximate solution for x is about 0.5496. This gives us a practical understanding of the value that satisfies the original exponential equation.

Why Use Logarithms?

You might be wondering, “Why go through all this logarithm business?” Great question! Logarithms are essential for solving exponential equations where the variable is in the exponent. Without logarithms, it would be nearly impossible to isolate x in equations like $2^{15x+5} = 12^{14x-4}$.

The logarithm property $\ln(a^b) = b \ln(a)$ is the magic behind this method. It allows us to bring the exponent down as a coefficient, transforming the exponential equation into a linear equation that we can solve using basic algebra. This transformation is the cornerstone of solving many exponential problems.

Common Mistakes to Avoid

When working with exponential equations and logarithms, it’s easy to make a few common mistakes. Let’s highlight some of these so you can steer clear of them:

1. Incorrectly Applying Logarithm Properties: Make sure you correctly apply logarithm properties, especially the power rule. A common mistake is to misapply or forget the rule $\ln(a^b) = b \ln(a)$.

2. Arithmetic Errors: Pay close attention to the arithmetic, especially when expanding and rearranging the equation. It’s easy to make a mistake when dealing with multiple terms and negative signs.

3. Rounding Too Early: If you’re approximating the solution, avoid rounding intermediate values. Round only at the final step to maintain accuracy. Rounding too early can lead to significant errors in your final answer.

4. Forgetting the Exact Solution: Remember that the exact solution is crucial. The calculator approximation is useful for understanding the value, but the exact solution provides the most accurate representation.

Practice Makes Perfect

The best way to master solving exponential equations is through practice. Try solving similar equations on your own, and you’ll quickly become more comfortable with the process. Here are a few practice problems to get you started:

1. $3^{2x+1} = 5^{x-2}$
2. $4^{3x-2} = 7^{2x+1}$
3. $9^{x+3} = 15^{2x-1}$

Work through these problems step by step, and you’ll build your skills and confidence.

Conclusion

So, guys, that's how you solve exponential equations using logarithms! By understanding the properties of logarithms and following a step-by-step approach, you can tackle even the trickiest exponential equations. Remember to express your solution in its exact form and use a calculator to get a practical approximation. Keep practicing, and you’ll become an exponential equation-solving pro in no time!

Solving exponential equations using logarithms is a powerful technique in mathematics. Whether you are dealing with complex equations in calculus or simple problems in algebra, mastering this skill is crucial. Remember, the key is to apply logarithms correctly, manipulate the equations carefully, and always double-check your work. Happy solving!