Solving Exponential & Logarithmic Equations Algebraically

Hey math enthusiasts! Today, we're diving deep into the fascinating world of exponential and logarithmic equations. If you've ever felt a little lost when trying to solve these algebraically, you're in the right place. We're going to break down several examples, rounding our answers to three decimal places to keep things precise. So, grab your calculators, and let's get started!

Equation 36: $3 e ^{-5 x }=132$

When we approach exponential equations, our primary goal is to isolate the exponential term. In this case, our exponential term is $e^{-5x}$. So, how do we get it by itself?

First things first, let's divide both sides of the equation by 3. This gives us:

$e^{-5x} = \frac{132}{3} = 44$

Now, we have $e^{-5x} = 44$. To get rid of that pesky exponent and solve for $x$, we need to use the natural logarithm (ln). Remember, the natural logarithm is the inverse operation of the exponential function with base e. Applying the natural logarithm to both sides, we get:

$\ln(e^{-5x}) = \ln(44)$

The beauty of logarithms is that they allow us to bring the exponent down as a coefficient. So, $-5x$ comes down in front:

$-5x \ln(e) = \ln(44)$

Since $\ln(e)$ is just 1, our equation simplifies to:

$-5x = \ln(44)$

Now, it’s just a matter of isolating $x$ by dividing both sides by -5:

$x = \frac{\ln(44)}{-5}$

Using a calculator, we find:

$x \approx -0.757$

So, the solution to the equation $3e^{-5x} = 132$, rounded to three decimal places, is approximately -0.757. Remember folks, the key here was isolating the exponential term and then using the natural logarithm to solve for the variable. This approach will be our trusty companion as we tackle more exponential equations. Let’s move on to our next example!

Equation 37: $2^x+13=35$

In this equation, we're dealing with an exponential term that has a base of 2. The same principle applies here: we want to isolate the exponential part, which is $2^x$.

So, our first step is to subtract 13 from both sides of the equation. This gives us:

$2^x = 35 - 13 = 22$

Now we have $2^x = 22$. To solve for $x$, we need to get rid of the exponent. Since our base is 2, we can use logarithms to help us. Specifically, we'll use the logarithm with base 2, often written as $\log_2$. Applying this to both sides, we get:

$\log_2(2^x) = \log_2(22)$

Just like with the natural logarithm, the exponent $x$ can now be brought down as a coefficient:

$x \log_2(2) = \log_2(22)$

Since $\log_2(2)$ is equal to 1, the equation simplifies to:

$x = \log_2(22)$

If your calculator doesn't have a base 2 logarithm function, don't worry! We can use the change of base formula, which states:

$\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$

Where b is the original base, a is the number we're taking the logarithm of, and c is the new base. We can use the common logarithm (base 10) or the natural logarithm (base e) for our calculations. Let's use the natural logarithm:

$x = \frac{\ln(22)}{\ln(2)}$

Using a calculator, we find:

$x \approx 4.459$

So, the solution to the equation $2^x + 13 = 35$, rounded to three decimal places, is approximately 4.459. Remember, isolating the exponential term and using logarithms (with the appropriate base or the change of base formula) are your best friends when solving these types of equations.

Equation 38: $-4(5^x)=-68$

Alright, let's tackle another exponential equation! In this case, we have $-4(5^x) = -68$. Remember our strategy? That's right, we need to isolate the exponential term, which is $5^x$ in this instance.

To get $5^x$ by itself, our first step is to divide both sides of the equation by -4:

$5^x = \frac{-68}{-4} = 17$

Now we're looking at $5^x = 17$. Just like in the previous example, we need to use logarithms to solve for $x$. Since the base of our exponential term is 5, we could use $\log_5$, but let's stick with the change of base formula to keep things consistent and use the natural logarithm. Applying the natural logarithm to both sides, we get:

$\ln(5^x) = \ln(17)$

Bring down the exponent as a coefficient:

$x \ln(5) = \ln(17)$

Isolate $x$ by dividing both sides by $\ln(5)$:

$x = \frac{\ln(17)}{\ln(5)}$

Time to punch those numbers into a calculator:

$x \approx 1.760$

Therefore, the solution to the equation $-4(5^x) = -68$, rounded to three decimal places, is approximately 1.760. You guys are getting the hang of this, right? Isolating the exponential term and using logarithms—that’s the winning formula!

Equation 39: $2 e^{x-3}-1=4$

Okay, let's keep this momentum going with another exciting exponential equation: $2e^{x-3} - 1 = 4$. You know the drill by now, right? Our mission is to isolate the exponential term, which in this case is $e^{x-3}$.

First, let's add 1 to both sides of the equation:

$2e^{x-3} = 4 + 1 = 5$

Now, divide both sides by 2:

$e^{x-3} = \frac{5}{2} = 2.5$

Fantastic! We've got $e^{x-3} = 2.5$. Since we have the exponential function with base e, we're going to use the natural logarithm. Apply it to both sides:

$\ln(e^{x-3}) = \ln(2.5)$

The exponent comes down in front:

$(x-3)\ln(e) = \ln(2.5)$

Remember, $\ln(e)$ is just 1, so we have:

$x - 3 = \ln(2.5)$

Now, let's isolate $x$ by adding 3 to both sides:

$x = \ln(2.5) + 3$

Grab your calculators, folks:

$x \approx 3.916$

So, the solution to the equation $2e^{x-3} - 1 = 4$, rounded to three decimal places, is approximately 3.916. Keep practicing these steps, and you’ll be solving exponential equations like a pro in no time!

Equation 40: $e^{2 x}-7 e^x+10=0$

Now, let's tackle a slightly different beast: $e^{2x} - 7e^x + 10 = 0$. This equation looks a bit more complex, but don't worry, we can handle it! Notice that this equation has a form that's similar to a quadratic equation. To make this more apparent, let's use a substitution.

Let $y = e^x$. Then, $e^{2x}$ can be written as $(e^x)^2$, which is just $y^2$. Our equation now becomes:

$y^2 - 7y + 10 = 0$

Ah, a quadratic equation! We can factor this equation quite nicely:

$(y - 5)(y - 2) = 0$

This gives us two possible values for $y$:

$y = 5$ or $y = 2$

But remember, we're not trying to solve for $y$; we want to find $x$. So, we need to substitute back $e^x$ for $y$:

$e^x = 5$ or $e^x = 2$

Now we have two separate exponential equations to solve. Let's start with $e^x = 5$. We'll use the natural logarithm:

$\ln(e^x) = \ln(5)$

$x = \ln(5)$

Using a calculator:

$x \approx 1.609$

Now, let's solve $e^x = 2$:

$\ln(e^x) = \ln(2)$

$x = \ln(2)$

Using a calculator:

$x \approx 0.693$

So, the solutions to the equation $e^{2x} - 7e^x + 10 = 0$, rounded to three decimal places, are approximately 1.609 and 0.693. See how that substitution helped us turn a tricky equation into something much more manageable? Keep that trick in your mathematical toolkit!

Equation 41: $x e^x+e^x=0$

Alright, let's dive into equation 41: $xe^x + e^x = 0$. This one looks interesting! We've got $x$ both inside and outside the exponential term, so we need to be strategic in our approach.

The first thing to notice here is that both terms in the equation have a common factor of $e^x$. Let's factor that out:

$e^x(x + 1) = 0$

Now, we have a product of two factors that equals zero. This means that either $e^x = 0$ or $x + 1 = 0$. Let's consider each case.

First, let's think about $e^x = 0$. Is there any value of $x$ that would make this true? Remember, the exponential function $e^x$ always produces a positive value, no matter what $x$ is. It never actually reaches zero. So, this part of the equation doesn't give us a solution.

Now, let's look at the second factor: $x + 1 = 0$. This one is straightforward to solve:

$x = -1$

So, the solution to the equation $xe^x + e^x = 0$ is simply -1. No rounding needed here! Isn’t it cool how factoring can simplify even complex-looking equations? Always be on the lookout for common factors, guys—they can be lifesavers.

Equation 42: $\ln 3 x=6.4$

Let’s move on to equation 42: $\ln(3x) = 6.4$. Ah, a logarithmic equation! When we're solving these, we need to remember that logarithms are the inverse of exponential functions. In this case, we have the natural logarithm, which is the logarithm with base e.

To get rid of the logarithm, we can exponentiate both sides of the equation using the base e. This means we raise e to the power of both sides:

$e^{\ln(3x)} = e^{6.4}$

The magic of inverse functions happens on the left side: $e^{\ln(3x)}$ simplifies to just $3x$:

$3x = e^{6.4}$

Now, we just need to isolate $x$ by dividing both sides by 3:

$x = \frac{e^{6.4}}{3}$

Time for the calculator:

$x \approx 200.277$

So, the solution to the equation $\ln(3x) = 6.4$, rounded to three decimal places, is approximately 200.277. Remember, when dealing with logarithmic equations, exponentiating is your key move to unlock the variable. You're doing great, guys!

By working through these examples step by step, I hope you've gained a solid understanding of how to solve exponential and logarithmic equations algebraically. The key takeaways here are:

• Isolate the exponential or logarithmic term.
• Use logarithms to solve exponential equations and exponentiation to solve logarithmic equations.
• Don't forget the change of base formula if you need to use a different base for your logarithms.
• Look for opportunities to factor or use substitution to simplify the equation.

Keep practicing, and these types of problems will become second nature. Happy solving, mathletes!