Finding Solutions: Which Equations Work With X=2?

Hey math enthusiasts! Let's dive into some equations and figure out which ones have a solution of x = 2. It's like a treasure hunt where we're testing each equation to see if it holds true when we plug in the value 2 for x. We'll break down each equation step-by-step, making sure we understand how to approach these kinds of problems. This is a great way to brush up on your algebra skills and get a deeper understanding of how equations work. Ready to get started? Let's go!

Equation 1: $\log _2(5 x+6)=4$

Alright, let's start with our first equation: $\log _2(5x + 6) = 4$. This is a logarithmic equation, and to solve it, we need to understand the relationship between logarithms and exponents. Remember, the logarithm tells us the exponent to which we must raise the base to get a certain number. In this case, the base is 2, and the logarithm equals 4. So, we can rewrite this equation in exponential form. This means that $2^4$ should equal $(5x + 6)$. Let's calculate that real quick. $2^4 = 16$. So, our equation now looks like this: $5x + 6 = 16$. Now, we're dealing with a simple linear equation, which is super easy to solve. Our goal here is to isolate x to find its value. First, subtract 6 from both sides of the equation. This gives us: $5x = 10$. Finally, to isolate x, divide both sides by 5. This gets us $x = 2$. Bingo! The first equation has x = 2 as its solution. That was pretty straightforward, right? We transformed the logarithmic equation into an exponential one and then solved the resulting linear equation. Keep in mind that understanding the properties of logarithms is key to tackling these types of problems. You can always double-check by substituting the solution back into the original equation to ensure it is correct.

So, when we substitute x = 2 back into our original equation, $\log _2(5x + 6) = 4$, we get $\log _2(5(2) + 6) = 4$. This simplifies to $\log _2(10 + 6) = 4$, which further simplifies to $\log _2(16) = 4$. And since $2^4 = 16$, the equation holds true. This confirms that x = 2 is indeed a solution to this equation. Awesome!

Equation 2: $\log _x 16=4$

Okay, let's move on to the second equation: $\log _x 16 = 4$. This one's a little different because the variable x is the base of the logarithm. But don't sweat it; the process is still similar. We need to rewrite this logarithmic equation in exponential form. Remember, $\log_b a = c$ can be rewritten as $b^c = a$. Applying this, our equation $\log_x 16 = 4$ becomes $x^4 = 16$. Our goal is to find the value of x that, when raised to the power of 4, equals 16. To solve for x, we can take the fourth root of both sides of the equation. In math terms, this is written as $\sqrt[4]{x^4} = \sqrt[4]{16}$. The fourth root of $x^4$ is simply x, and the fourth root of 16 is 2, because $2^4 = 16$. So, $x = 2$ is our solution! Isn't it cool how everything connects? By converting the logarithmic form into exponential form, we were able to isolate x and solve the equation. The key to solving such problems lies in understanding the definitions of logarithms and exponents and how to convert between them. Being able to recognize and apply these concepts is crucial. Don't forget that after you find a solution, it's always smart to double-check it by substituting it back into the original equation. Let’s do that to confirm our answer.

To double-check if x = 2 is correct, let's substitute it back into the original equation: $\log_x 16 = 4$. So, we have $\log_2 16 = 4$. Does this hold true? Yes, because $2^4 = 16$. Therefore, x = 2 is indeed a valid solution for this second equation. Great job! Two down, and two to go. We're on a roll.

Equation 3: $\log _3(6 x+4)=3$

Alright, let's keep the ball rolling with our third equation: $\log_3(6x + 4) = 3$. This is another logarithmic equation, and just like before, we're going to transform it into its exponential form to solve it. In this case, the base is 3, and we have $\log_3(6x + 4) = 3$. Remember, this means $3^3$ should equal $6x + 4$. Calculating $3^3$, we get $3 * 3 * 3 = 27$. So, our equation now becomes $6x + 4 = 27$. We need to isolate x to find its value. First, subtract 4 from both sides of the equation. This gives us: $6x = 23$. Now, divide both sides by 6 to isolate x: $x = \frac{23}{6}$. So, the solution to this equation is $x = \frac{23}{6}$. Notice that this is not equal to 2. Therefore, this equation does not have x = 2 as a solution. In solving this equation, we again saw the importance of knowing how to convert a logarithmic equation into an exponential one. This is a fundamental skill that unlocks your ability to solve a wide variety of logarithmic problems. This transformation simplifies the equation, making it easier to isolate the variable and find its value. Always remember to check your work; however, in this case, we have confirmed that our original hypothesis is wrong. We are looking for an x that equals 2, and that is not what we got here.

Let's do a quick double-check just to be sure. If we plug $x = \frac{23}{6}$ into our original equation, $\log_3(6x + 4) = 3$, we get $\log_3(6(\frac{23}{6}) + 4) = 3$, which simplifies to $\log_3(23 + 4) = 3$, or $\log_3(27) = 3$. Since $3^3 = 27$, the equation holds true. This confirms that $x = \frac{23}{6}$ is indeed the correct solution for this equation. But as previously stated, it is not equal to 2, so this is not one of our solutions.

Equation 4: $\log _x 36=6$

Finally, let's tackle the last equation: $\log_x 36 = 6$. Just like the second equation, the variable x is the base of the logarithm here. Let's convert this into exponential form. We have $x^6 = 36$. To find the value of x, we need to take the sixth root of both sides of the equation. In mathematical terms, this is written as $\sqrt[6]{x^6} = \sqrt[6]{36}$. The sixth root of $x^6$ is simply x. The sixth root of 36 is not a whole number; however, we can simplify this, or we can understand that it is not equal to 2. Remember, $2^6 = 64$. So, our solution is not 2. To get the exact value, we can use a calculator to find that $x = \sqrt[6]{36}$, which is approximately 1.82. Thus, the solution to this equation is not 2. Remember, the ability to convert logarithmic equations into exponential form is key to solving these types of problems. This skill will help you simplify the equation, allowing you to isolate the variable and find its value. That is why it is extremely important to be familiar with the properties of logarithms and exponents. Being able to recognize and apply these concepts is crucial for solving these types of equations. Let's do our final double-check here as well.

Let’s double-check by substituting our approximate value back into the original equation, $\log_x 36 = 6$. So, we have $\log_{1.82} 36 = 6$. When you put this in a calculator, it confirms the answer. We know that x = 2 is not a solution to this equation. We have successfully worked through all four equations and determined which ones have x = 2 as the solution. Nicely done!

Conclusion: Which Equations Have x=2 as the Solution?

So, which equations did we find that have x = 2 as a solution? Let's recap. After working through all four equations, we found that:

• $\log _2(5 x+6)=4$ has x = 2 as a solution.
• $\log _x 16=4$ has x = 2 as a solution.

That's it, guys! We successfully identified which of the given equations have x = 2 as their solution. Great work everyone. Remember, practice makes perfect when it comes to solving equations. The more you work through these problems, the more comfortable and confident you'll become. Keep up the awesome work, and keep exploring the fascinating world of mathematics!