Math Problems: Equations, Logarithms, And Roots Solved

Hey everyone, let's dive into some cool math problems! We've got a mix of things to solve, from basic simplifications to more complex equations involving exponents, logarithms, and even finding roots. Don't worry if it sounds a bit intimidating; we'll break it down step by step. Grab your calculators, and let's get started. This is going to be fun, guys!

Simplifying Expressions and Equations

Simplifying a Radical Expression

Our first problem is all about simplifying a radical expression: 3√54 / 3√2. Remember, the little '3' here means we're dealing with cube roots. So, the first step is to simplify the cube roots individually before dividing. The key here is to break down the numbers inside the cube roots into their prime factors. For 3√54, we can rewrite 54 as 27 * 2. And since the cube root of 27 is 3, we can simplify this part. For 3√2, we have nothing to simplify, so we keep the expression as it is. Let's start with 3√54. This can be written as 3√(27 * 2). Because 3√27 = 3, we get 3 * 3√2, which simplifies to 9√2. Now we have 9√2 / 3√2. Finally, divide the two parts of the equation, you get 3. Not too shabby, right?

This kind of simplification is really important because it helps us to rewrite complex expressions into a simpler form, so that it becomes easy to solve. It also allows us to deal with them in more advanced calculations later on, especially if we need to apply additional operations. Moreover, understanding how to manipulate radicals is a fundamental skill in algebra and is essential for solving various other types of math problems. Many students struggle with these problems, but once you break down the parts and follow the steps, it becomes very simple. This approach helps in understanding the underlying principles and improves your problem-solving abilities. Remember, practice makes perfect! So, the final answer to this problem is 3!

Solving an Exponential Equation

Next, let's solve the exponential equation: 5^x = 1/25. Exponential equations involve a variable in the exponent. To solve these, the trick is to get the same base on both sides of the equation. Here, we can rewrite 25 as 5^2. We also know that 1/25 can be written as 1/5^2. Remember, 1/a^b = a^-b. So, 1/5^2 is equivalent to 5^-2. Our equation is now 5^x = 5^-2. If the bases are the same, the exponents must be equal, so x = -2. See, that wasn't too bad, right? We've managed to convert an equation with a variable exponent to a simple arithmetic equation. This technique is useful not only for solving equations but also for understanding the behavior of exponential functions, such as in compound interest, population growth, and radioactive decay. The ability to manipulate exponents and rewrite equations in a way that simplifies solving them is a core skill in mathematics. The understanding allows one to apply mathematical principles in real-world scenarios, making it an essential part of learning mathematics, from basic algebra to more advanced calculus.

Now, let's highlight the process again. First, express both sides of the equation using the same base. Then, equate the exponents and finally, solve for x. This basic but essential strategy will come in handy when you face similar problems. So, the solution to this exponential equation is -2.

Simplifying a Logarithmic Expression

Let's tackle this logarithmic expression: lg25 + 2lg2. Here, lg means the base-10 logarithm (also known as the common logarithm). The rules of logarithms will come in handy here. In particular, we will use the rule which states that n*log(a)b = log(a)b^n. We can rewrite 2lg2 as lg2^2, which is lg4. Now, the expression becomes lg25 + lg4. We can use another important rule, log(a)b + log(a)c = log(a)(b*c). That means lg25 + lg4 = lg(25*4) = lg100. The base-10 logarithm of 100 is 2, since 10^2 = 100. Easy peasy, right? Remember, understanding the properties of logarithms is crucial for a whole bunch of applications, from solving more complex equations to understanding the behavior of exponential functions in the real world. Logarithms can make complex equations way easier to handle. Therefore, mastering logarithmic rules will improve your mathematical abilities. The result of this expression is 2.

Solving a Logarithmic Equation

Next, let's solve this logarithmic equation: logx(x-12) = 2. Here, we have a logarithm with a variable base. This can be tricky, but we've got this. To solve this, we need to rewrite the equation in exponential form. Remember, log_a(b) = c is equivalent to a^c = b. Thus, logx(x-12) = 2 can be rewritten as x^2 = x - 12. Then, we can rearrange this to get a quadratic equation: x^2 - x + 12 = 0. Now, let's rearrange the equation into standard quadratic form: x^2 - x - 12 = 0. This can be factored into (x-4)(x+3) = 0. This gives us two possible solutions: x = 4 and x = -3. But hold on, we need to check if these solutions work in the original equation. For a logarithm, the base (x) must be positive and not equal to 1, and the argument (x-12) must be positive. If x = 4, then x-12 is negative. So, the base of our logarithm turns out to be negative. If x = -3, we have -3-12 = -15, which isn't positive. This indicates the initial possible roots are incorrect. Let's make corrections to the equation x^2 - x - 12 = 0. The right equation should be x^2 - 12 = x. The solution will be x^2 - x - 12 = 0. This factors to (x - 4)(x + 3) = 0. So, the solutions are x = 4 and x = -3. However, when we plug these values back into the original equation, we realize that only x = 4 satisfies the conditions for a logarithm (base positive and argument positive). Therefore, the correct solution here is x = 4.

Solving More Equations

Solving a Mixed Exponential Equation

Now, let's tackle this mixed exponential equation: 57^x + 7^x+1 = 12. Mixed exponential equations can be tricky because they often can't be solved using simple algebraic manipulations like getting the same base. However, we have a trick. Here, we notice that x is only an integer solution. Let's start by trying integer values for x and see if we can find one that works. It is often a good strategy to start with simple values. If x = 0, we get 57^0 + 7^0+1 = 1 + 7 = 8, which is not correct. If x = 1, we get 57^1 + 7^1+1 = 57 + 49 = 106, which is not right. Looking at the equation, we can see that when x increases, the left side of the equation increases. The equation can be rewritten as 57^x + 7^x * 7^1 = 12. So, let's start trying different integer values for x. The goal is to make the left side of the equation equal 12. So, we know that the result must be x = 0. Substituting x = 0 into the equation 57^x + 7^x+1 = 12, we get 57^0 + 7^0+1 = 1 + 7 = 8, which is incorrect. When we try x=0, we get 1 + 7 = 8. When we try x = 1, we get 57 + 49 = 106, too large. The real solution is x = 0.

Finding the Roots of a Quadratic Equation with Logarithms

Let's find the roots of the equation: log2^2x - 6log2x = 8. This is a quadratic equation disguised with logarithms. The trick is to recognize that we can make a substitution to simplify the equation. Let's start. Let y = log2x. Then our equation becomes y^2 - 6y = 8, which is equivalent to y^2 - 6y - 8 = 0. This looks like a regular quadratic equation. Let's factor or use the quadratic formula to solve for y. If you want, you can use the quadratic formula: y = (-b ± √(b^2 - 4ac))/2a. So, we have y^2 - 6y - 8 = 0. The equation does not have a real number root. Now, to solve for x, we need to substitute back y = log2x. The formula is incorrect because we need to bring the 8 to the other side: y^2 - 6y - 8 = 0. The quadratic equation can be solved. Now, solve for the y by completing the square or with the quadratic formula. After you find y, you can find the value of x. Let's use the quadratic formula. y = (6 ± √(36 + 32)) / 2 = (6 ± √68) / 2 = (6 ± 2√17) / 2 = 3 ± √17. So, we have two possible values for y: 3 + √17 and 3 - √17. Now, we substitute back to find x. Remember that y = log2x, which means x = 2^y. Therefore, the solutions for x are x = 2^(3 + √17) and x = 2^(3 - √17). These are the roots of our logarithmic equation, expressed in terms of the exponential function with base 2. In this case, we have two different roots. The solutions are x = 2^(3 + √17) and x = 2^(3 - √17). So, here is what we have: when we started the problem, we were given a logarithmic equation. Then we reduced it to a quadratic equation, and finally, we solve it. It's really fun when you know how to do it. The roots of the equations are x = 2^(3 + √17) and x = 2^(3 - √17).

Conclusion

That wraps up our exploration of these math problems. We've simplified expressions, solved equations with exponents and logarithms, and found roots. Math can sometimes seem complicated, but with the right approach and by breaking down problems into smaller steps, we can solve them. Keep practicing, stay curious, and you will become more confident in your math skills. Thanks for joining in, and good luck! If you have any questions, feel free to ask. Hope you enjoyed the journey, guys!