Unlocking Math: Equations & Logarithms Explained

Hey Plastik Magazine readers! Ready to dive into some math problems? Don't worry, we'll break it down so it's super easy to understand. We're gonna tackle some equations and logarithm questions, step by step. Let's get started, shall we?

Question 21: Solving Exponential Equations

Alright, let's start with question 21: Solve for x in the equation $3^{2x} - 3^{x+2} = 3^{x+1} - 27$. This might look a little intimidating at first, but trust me, we can handle it. The key here is to simplify and manipulate the equation to isolate x. This involves understanding the properties of exponents and a little bit of algebraic manipulation. We'll break down the steps to find the solution for x and then check which of the provided options (A, B, C, or D) is the correct answer. The good thing is that we're going to use the power of exponents. This means that we are going to need to simplify the exponents as much as possible.

First, let's rewrite the equation, making sure to apply the exponent rules. Remember that $3^{x+2}$ is the same as $3^x * 3^2$ and that $3^{x+1}$ is equivalent to $3^x * 3^1$. Let's start transforming and rewriting our original problem as the first step to get to the solution. The most important step in all these kinds of problems is to learn how to manipulate the original expression until it is simple enough. Now, let’s rewrite the equation using these rules. This gives us: $3^{2x} - (3^x * 3^2) = (3^x * 3^1) - 27$. Now that we've expanded the exponents, let's simplify further. We know that $3^2 = 9$ and $3^1 = 3$. Our equation now looks like this: $3^{2x} - 9 * 3^x = 3 * 3^x - 27$. This is already looking much better, right? We can now work on simplifying and moving the terms around to get our equation into a more manageable form. After that, we must make sure we can isolate the x variable, which is what we want. The next step is to get all the terms involving $3^x$ on one side and the constants on the other side. Let's move the terms around, shall we? You know we want to get the same base, which will simplify the math a lot.

Now, let's make a substitution to make things even clearer. Let $y = 3^x$. Then, $3^{2x}$ becomes $y^2$. Our equation now looks like this: $y^2 - 9y = 3y - 27$. This looks much more familiar now, doesn't it? Let’s put all the terms on one side to get a quadratic equation: $y^2 - 12y + 27 = 0$. This is now a standard quadratic equation. We can solve it by factoring. Finding two numbers that multiply to 27 and add up to -12, we can easily factor the equation. Those numbers are -3 and -9. We can now factor our quadratic equation to: $(y - 3)(y - 9) = 0$. This gives us two possible values for y: $y = 3$ or $y = 9$. Remember that $y = 3^x$. Let's substitute back to find the values of x. Let's first solve for $y=3$, the solution will be $3^x = 3$. Since the bases are the same, the exponents must be equal, so $x = 1$. Now, let's solve for $y=9$, the solution will be $3^x = 9$. Since 9 is $3^2$, we have $3^x = 3^2$. Therefore, $x = 2$. So, the solutions for x are 1 and 2. Let's go back to our answer options. Checking our answer options, we can see that the correct option is B: 1 or 2.

Question 22: Logarithm Manipulation

Okay, guys, let's move on to question 22. This one deals with logarithms: If $\log_3 a - 2 = 3 \log_3 b$, express a in terms of b. This problem tests our understanding of logarithm properties, such as the power rule and the change of base formula. The goal is to isolate a on one side of the equation and express it in terms of b. We will need to use our logarithm skills and knowledge to isolate the variable a. Let's start with the given equation: $\log_3 a - 2 = 3 \log_3 b$. First, we need to rewrite the equation to isolate the terms. Let's get the terms involving logarithms on one side and the constant on the other. First, let's add 2 to both sides of the equation. This gives us $\log_3 a = 3 \log_3 b + 2$. Now that we've isolated $\log_3 a$, we can start simplifying the right side of the equation. We want to combine the logarithmic terms as much as possible. This way, we will get the variable a much faster.

We can use the power rule of logarithms, which states that $n \log_c m = \log_c m^n$. Applying this rule to the term $3 \log_3 b$, we get $\log_3 b^3$. Our equation now becomes: $\log_3 a = \log_3 b^3 + 2$. Now, let’s deal with the constant term. Remember that $2$ can be expressed as $2 \log_3 3$. Using the power rule again, we can rewrite $2 \log_3 3$ as $\log_3 3^2$, which simplifies to $\log_3 9$. Now we must combine the terms. Substituting this back into our equation, we get: $\log_3 a = \log_3 b^3 + \log_3 9$. Now that we have two logarithmic terms on the right side, we can use the product rule of logarithms, which states that $\log_c m + \log_c n = \log_c (m * n)$. This allows us to combine the terms. Applying the product rule, we have: $\log_3 a = \log_3 (9b^3)$. Since the logarithms have the same base, we can equate the arguments. Therefore, $a = 9b^3$. Now, we must check the answer options to get the final solution. The correct answer option is C: $a = 9b^3$.

Question 23: Solving Logarithmic Equations

Let’s tackle question 23: Solve for x in the equation $\log_2(12x - 10) = 1 + \log_2(4x - 10)$. This problem combines our knowledge of logarithms and equations. The key is to simplify the equation using logarithm properties and then solve for x. Remember that we need to isolate the variable, which is what we need to get to the answer. The goal is to isolate x and find the solution. Let's start by rewriting the equation and making sure we apply our logarithm rules to it. Let's simplify the equation. First, we need to get all the logarithmic terms on one side of the equation. We can do this by subtracting $\log_2(4x - 10)$ from both sides. Doing this gives us: $\log_2(12x - 10) - \log_2(4x - 10) = 1$. The next step involves using the quotient rule of logarithms. This rule states that $\log_c m - \log_c n = \log_c \frac{m}{n}$. Applying this rule, we can combine the logarithms on the left side: $\log_2 \frac{(12x - 10)}{(4x - 10)} = 1$. Now that we have a single logarithm, we can rewrite the equation in exponential form. Remember that $\log_c m = n$ is equivalent to $c^n = m$. Converting our logarithmic equation into exponential form, we get $2^1 = \frac{(12x - 10)}{(4x - 10)}$. This simplifies to $2 = \frac{(12x - 10)}{(4x - 10)}$.

Now, let's solve for x. Multiply both sides by $(4x - 10)$ to eliminate the fraction: $2(4x - 10) = 12x - 10$. This simplifies to $8x - 20 = 12x - 10$. Next, we need to isolate x. Let's subtract $8x$ from both sides: $-20 = 4x - 10$. Now, add 10 to both sides: $-10 = 4x$. Finally, divide both sides by 4 to solve for x: $x = -\frac{10}{4}$, which simplifies to $x = -\frac{5}{2}$ or $x = -2.5$. Before we celebrate, we must check our solution. We must ensure our solution doesn't make the arguments of the logarithms negative. Let’s plug $x = -2.5$ back into the original equation: $\log_2(12(-2.5) - 10) = 1 + \log_2(4(-2.5) - 10)$. Let's simplify the arguments of the logarithms. This will give us $\log_2(-30 - 10)$ and $\log_2(-10 - 10)$. Since we can't take the logarithm of a negative number, the solution $x = -2.5$ is not valid. Therefore, there is no solution to this equation. We must remember to check the solution in the original equation to ensure that there are no negative logarithms. Since this solution doesn’t work, we can say that there is no solution for the given equation.