Solving Exponential Equations: A Step-by-Step Guide

Hey guys! Let's dive into the exciting world of exponential equations. We're going to break down several examples step-by-step, so you can conquer these problems with confidence. Grab your pencils, and let's get started!

a) 3^(x+3) = 81

Let's tackle our first exponential equation: 3^(x+3) = 81. The key here is to express both sides of the equation with the same base. Since 81 is a power of 3 (specifically, 3^4), we can rewrite the equation as:

3^(x+3) = 3^4

Now that the bases are the same, we can equate the exponents:

x + 3 = 4

Solving for x, we subtract 3 from both sides:

x = 4 - 3

x = 1

So, the solution to the equation 3^(x+3) = 81 is x = 1. To make sure this is clear, remember that the goal is always to get both sides of the equation to have the same base. Once you've achieved that, just set the exponents equal to each other and solve for the variable. This approach simplifies what initially seems like a complex problem. Always double-check your answer by plugging it back into the original equation to confirm it holds true. Understanding this fundamental technique will give you a strong foundation for tackling more complex exponential equations down the road. Keep practicing, and soon you'll be solving these equations like a pro!

b) 8^(3x+1) = 64

Next up, we have 8^(3x+1) = 64. Again, we need to express both sides with the same base. Notice that both 8 and 64 are powers of 2 (8 = 2^3 and 64 = 2^6). Alternatively, you can also express 64 as a power of 8 (64 = 8^2). Let's use the base 8 for simplicity:

8^(3x+1) = 8^2

Now, equate the exponents:

3x + 1 = 2

Subtract 1 from both sides:

3x = 2 - 1

3x = 1

Divide by 3:

x = 1/3

Therefore, the solution to the equation 8^(3x+1) = 64 is x = 1/3. Keep in mind, expressing both sides with the same base is the golden rule here. Sometimes, it may not be immediately obvious what the common base is, but with a little bit of manipulation, you can usually find it. Also, remember that there's often more than one way to solve these problems. For example, we could have converted both 8 and 64 to base 2. The important thing is to choose the approach that makes the most sense to you and minimizes the chance of making errors. Checking your work by substituting the solution back into the original equation is a great habit to develop, ensuring accuracy every time.

c) 8^(1-2x) = √√27

This one looks a bit trickier: 8^(1-2x) = √√27. First, let's simplify the right side of the equation. Remember that √√27 means the fourth root of 27, which can be written as 27^(1/4). Also, 27 is 3^3, so we can rewrite the right side as (3³⁾(1/4) = 3^(3/4).

Now, let's express both sides with the base 2 and 3 respectively. Since 8 = 2^3, the left side becomes (2³⁾(1-2x) = 2^(3-6x). Therefore, our equation is:

2^(3-6x) = 3^(3/4)

Oh no! We cannot express both sides with the same base. There must be a typo in the original equation. Assuming the equation was meant to be 8^(1-2x) = √√16, then we can proceed as follows:

8^(1-2x) = √√16

Simplify the right side: √√16 = √(4) = 2. Also, 8 = 2^3. So, the equation becomes:

(2³⁾(1-2x) = 2

2^(3-6x) = 2^1

Equate the exponents:

3 - 6x = 1

Subtract 3 from both sides:

-6x = -2

Divide by -6:

x = 1/3

So, if the equation was 8^(1-2x) = √√16, the solution is x = 1/3. Guys, always double-check the original problem statement to make sure you're solving the right equation! This example highlights the importance of careful reading and attention to detail. Even a small mistake in the problem can lead to a completely different solution. When you encounter difficulties, take a moment to review the problem statement and make sure you've correctly transcribed all the information. It can save you a lot of frustration in the long run!

d) √(16+1) = √64

Okay, let's look at √(16+1) = √64. Simplify both sides:

√17 = 8

This equation is not true, so there's no solution. It's a contradiction! Sometimes, you'll encounter equations that simply don't have a solution. This can happen for a variety of reasons, such as conflicting conditions or inherent impossibilities within the equation itself. In such cases, the correct answer is to state that there is no solution. Recognizing these situations is an important skill in mathematics. It prevents you from wasting time trying to find a solution that doesn't exist and encourages you to think critically about the nature of the equation itself. Always be prepared for the possibility of a "no solution" answer!

e) √√5^x * 25^(x+1) = (0.2)^(1-x)

Now, let's solve √√5^x * 25^(x+1) = (0.2)^(1-x). Rewrite the equation using fractional exponents and a common base of 5:

(5^(x/4)) * (5²⁾(x+1) = (5^(-1))(1-x)

Simplify:

5^(x/4) * 5^(2x+2) = 5^(x-1)

Combine the exponents on the left side:

5^(x/4 + 2x + 2) = 5^(x-1)

Equate the exponents:

x/4 + 2x + 2 = x - 1

Multiply everything by 4 to get rid of the fraction:

x + 8x + 8 = 4x - 4

Combine like terms:

9x + 8 = 4x - 4

Subtract 4x from both sides:

5x + 8 = -4

Subtract 8 from both sides:

5x = -12

Divide by 5:

x = -12/5

So, the solution is x = -12/5. This problem demonstrates the power of using fractional exponents and common bases to simplify complex equations. By breaking down each term into its fundamental components, we were able to combine exponents, solve for x, and arrive at the solution. Remember, practice is key! The more you work with these types of problems, the more comfortable you'll become with the techniques involved. Don't be afraid to make mistakes along the way – they're an essential part of the learning process. Just keep practicing, and you'll master these equations in no time!

f) 0.01^(100x) = 0.02

Let's solve 0.01^(100x) = 0.02. Rewrite the equation in terms of fractions:

(1/100)^(100x) = 2/100

Rewrite using exponents:

(10^(-2))(100x) = 2 * 10^(-2)

Simplify:

10^(-200x) = 2 * 10^(-2)

To solve for x, we'll use logarithms. Take the logarithm base 10 of both sides:

log10(10^(-200x)) = log10(2 * 10^(-2))

-200x = log10(2) - 2

Divide by -200:

x = (2 - log10(2)) / 200

x ≈ (2 - 0.301) / 200

x ≈ 1.699 / 200

x ≈ 0.008495

So, the solution is approximately x ≈ 0.008495. This problem introduces logarithms as a tool for solving exponential equations. When you can't easily express both sides with the same base, logarithms provide a powerful alternative. The key is to remember the properties of logarithms and how to apply them correctly. Practice using logarithms with different bases to become comfortable with this technique. And remember, calculators are your friends when dealing with logarithmic calculations!

g) (10^(x+2) * 100^(-3) )/ 100000000 + 3 = (125)^(x/3) / (0.4)^(2x-3)

This equation is a beast! (10^(x+2) * 100^(-3) )/ 100000000 + 3 = (125)^(x/3) / (0.4)^(2x-3). Let's simplify it step-by-step. First, rewrite everything in terms of powers of 2, 5, and 10:

(10^(x+2) * (10²⁾(-3)) / 10^8 + 3 = (5³⁾(x/3) / ((2/5))^(2x-3)

Simplify further:

(10^(x+2) * 10^(-6)) / 10^8 + 3 = 5^x / (2^(2x-3) / 5^(2x-3))

10^(x-4) / 10^8 + 3 = 5^x * 5^(3-2x) / 2^(2x-3)

10^(x-12) + 3 = 5^(3-x) / 2^(2x-3)

10^(x-12) + 3 = (5^3 / 5^x) / (2^(2x) / 2^3)

This equation is getting complicated, and it's not easy to isolate x algebraically. Numerical methods or computational tools might be necessary to find an approximate solution. Alternatively, there might be a typo in the original equation. Without further simplification or context, it's difficult to proceed analytically. Complex equations like this often require advanced techniques or numerical solvers to find solutions. Don't be discouraged if you can't solve them by hand – even experienced mathematicians sometimes rely on computational tools for such problems.

h) √√2^x * √√4^2 = √8^(x-6)

Let's solve √√2^x * √√4^2 = √8^(x-6). Rewrite the equation using fractional exponents:

(2^(x/4)) * (4^(2/4)) = (8^((x-6)/2))

Simplify:

2^(x/4) * 4^(1/2) = 8^((x-6)/2)

Since 4^(1/2) = 2, we have:

2^(x/4) * 2 = 8^((x-6)/2)

Rewrite everything with base 2:

2^(x/4) * 2^1 = (2³⁾((x-6)/2)

2^(x/4 + 1) = 2^(3(x-6)/2)

Equate the exponents:

x/4 + 1 = (3x - 18) / 2

Multiply by 4 to eliminate fractions:

x + 4 = 2(3x - 18)

x + 4 = 6x - 36

Subtract x from both sides:

4 = 5x - 36

Add 36 to both sides:

40 = 5x

Divide by 5:

x = 8

So, the solution is x = 8. This problem reinforces the importance of simplifying expressions and using fractional exponents effectively. By rewriting everything with a common base, we were able to equate the exponents and solve for x. Always remember to double-check your work and make sure your solution satisfies the original equation.

i) (1/4)^(x-6) * (3√(3^3 * x))^2 = (4/3)^x * 6√27

Finally, let's tackle (1/4)^(x-6) * (3√(3^3 * x))^2 = (4/3)^x * 6√27. This one is a doozy! It seems there might be a typo in the equation, particularly with the terms involving radicals and the variable x under the radical. Assuming a more simplified version of the equation, we can proceed as follows:

Let's assume the equation is: (1/4)^(x-6) * (3√(27))^2 = (4/3)^x * 6√27

Simplify the radicals:

√27 = √(3^3) = 3√3

So the equation becomes:

(1/4)^(x-6) * (3 * 3√3)^2 = (4/3)^x * 6 * 3√3

(1/4)^(x-6) * (9√3)^2 = (4/3)^x * 18√3

(1/4)^(x-6) * 81 * 3 = (4/3)^x * 18√3

(1/4)^(x-6) * 243 = (4/3)^x * 18√3

Rewrite with base 4/3:

(4/3)^(6-x) * 243 = (4/3)^x * 18√3

Divide both sides by 243:

(4/3)^(6-x) = (4/3)^x * (18√3) / 243

(4/3)^(6-x) = (4/3)^x * (2√3) / 27

(4/3)^(6-x) = (4/3)^x * (2 * 3^(1/2)) / (3^3)

(4/3)^(6-x) = (4/3)^x * 2 / (3^(5/2))

This equation is still complex. Without the correct original equation, we cannot find a simple algebraic solution for x. Numerical methods might be required. Remember, always double-check the original problem statement for any typos or errors. This can save you a lot of time and frustration when trying to solve complex equations!

Okay, guys, that was a marathon! We've tackled a variety of exponential equations, from simple to complex. Remember the key strategies: express both sides with the same base, use fractional exponents, and don't be afraid to use logarithms when necessary. Keep practicing, and you'll become an exponential equation master in no time!