Solve Logarithmic Equations: Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of logarithmic equations. These equations might seem intimidating at first, but with a little bit of algebraic finesse, we can crack them open and find their solutions. In this article, we will solve the logarithmic equation $\log x+2 \sqrt{\log x}-8=0$ step by step. We'll break down the process into manageable chunks, making it super easy to follow along. So, grab your thinking caps and let's get started!

Understanding Logarithmic Equations

Before we jump into solving our specific equation, let's take a moment to understand what logarithmic equations are all about. A logarithmic equation is simply an equation where the variable appears inside a logarithm. Remember, logarithms are the inverse of exponential functions. So, if we have an equation like $y = \log_b x$, it means that $b^y = x$. Understanding this relationship is crucial for solving logarithmic equations.

The key to solving logarithmic equations often involves manipulating the equation to isolate the logarithmic term and then converting it into its exponential form. This is where the properties of logarithms come in handy. For instance, we'll often use properties like $\log_b (mn) = \log_b m + \log_b n$ or $\log_b (m/n) = \log_b m - \log_b n$ to simplify the equation. However, in our particular problem, we'll use a slightly different approach involving substitution to make things easier. So, keep those logarithmic properties in the back of your mind, but let's focus on the specific technique we'll use for this equation.

Another important thing to remember when dealing with logarithmic equations is the domain of the logarithm. The argument of a logarithm (the thing inside the log) must always be positive. This means that if we find a solution for $x$, we need to check if it actually makes the argument of the logarithm positive. If it doesn't, we have to discard that solution. This step is crucial to avoid extraneous solutions. We'll see this in action when we solve our equation, so keep an eye out for it.

Step-by-Step Solution of $\log x+2 \sqrt{\log x}-8=0$

Okay, let's get down to business and solve the equation $\log x+2 \sqrt{\log x}-8=0$. This equation might look a bit tricky at first glance because of the square root and the logarithm. But don't worry, we'll tackle it systematically.

1. Substitution

The first trick we're going to use is substitution. This is a common technique for dealing with equations that have a repeating expression. Notice that we have $\log x$ and $\sqrt{\log x}$ in our equation. Let's make a substitution to simplify things. We'll let $y = \sqrt{\log x}$. This means that $y^2 = (\sqrt{\log x})^2 = \log x$. Now, we can rewrite our equation in terms of $y$. Substituting $y^2$ for $\log x$ and $y$ for $\sqrt{\log x}$, we get:

$y^2 + 2y - 8 = 0$

See how much simpler that looks? We've transformed a logarithmic equation into a good old quadratic equation. This is a huge step in the right direction!

2. Solve the Quadratic Equation

Now, we have a quadratic equation $y^2 + 2y - 8 = 0$. There are several ways to solve quadratic equations, such as factoring, completing the square, or using the quadratic formula. In this case, factoring is the easiest way to go. We need to find two numbers that multiply to -8 and add up to 2. Those numbers are 4 and -2. So, we can factor the quadratic equation as:

$(y + 4)(y - 2) = 0$

This gives us two possible solutions for $y$:

• $y + 4 = 0$ which means $y = -4$
• $y - 2 = 0$ which means $y = 2$

So, we have $y = -4$ and $y = 2$ as potential solutions. But remember, we're not looking for $y$, we're looking for $x$. We need to substitute back to find the values of $x$.

3. Substitute Back

Now comes the crucial step of substituting back to find the values of $x$. We know that $y = \sqrt{\log x}$. So, we have two cases to consider:

Case 1: $y = -4$

If $y = -4$, then we have $\sqrt{\log x} = -4$. Now, here's a crucial point: the square root of a number cannot be negative. The principal square root is always non-negative. Therefore, $\sqrt{\log x} = -4$ has no solution. This is a really important thing to keep in mind when solving equations involving square roots.

Case 2: $y = 2$

If $y = 2$, then we have $\sqrt{\log x} = 2$. To get rid of the square root, we can square both sides of the equation:

$(\sqrt{\log x})^2 = 2^2$

This simplifies to:

$\log x = 4$

Now, we have a simple logarithmic equation. Remember that if we don't write the base of the logarithm, it's assumed to be base 10. So, $\log x$ means $\log_{10} x$. To convert this logarithmic equation into exponential form, we use the definition of logarithms: if $\log_b x = y$, then $b^y = x$. In our case, we have $\log_{10} x = 4$, so:

$10^4 = x$

This gives us:

$x = 10000$

So, we have a potential solution: $x = 10000$.

4. Check the Solution

We're not quite done yet! We need to check if our solution is valid. Remember, the argument of a logarithm must be positive. In our original equation, we have $\log x$. So, we need to make sure that $x > 0$. Our solution $x = 10000$ is positive, so it satisfies this condition.

Now, let's plug $x = 10000$ back into the original equation to make sure it works:

$\log(10000) + 2\sqrt{\log(10000)} - 8 = 0$

We know that $\log(10000) = \log_{10}(10^4) = 4$. So, we have:

$4 + 2\sqrt{4} - 8 = 0$

$4 + 2(2) - 8 = 0$

$4 + 4 - 8 = 0$

$0 = 0$

This is true! So, $x = 10000$ is indeed a valid solution.

Final Answer

Therefore, the solution to the equation $\log x+2 \sqrt{\log x}-8=0$ is $x = 10000$. This corresponds to option A. {10,000}.

So, there you have it! We've successfully solved a logarithmic equation using substitution, factoring, and a bit of careful checking. I hope you found this step-by-step guide helpful. Remember, practice makes perfect, so keep solving those equations!

Final Answer: The final answer is $\boxed{{10,000}}$