Solving Logarithmic Equations: A Step-by-Step Guide

Hey guys! Today, we're diving into solving a logarithmic equation and verifying our solution using a graphing calculator. It might sound intimidating, but trust me, we'll break it down into easy-to-follow steps. Our equation is ${\log(5) + \log(-3x) = 2}$. Let's get started!

Step 1: Combine Logarithms

First things first, we need to combine the logarithms on the left side of the equation. Remember the logarithmic property that states ${\log(a) + \log(b) = \log(ab)}$. Applying this property, we get:

\begin{align*} \log(5) + \log(-3x) &= \log(5 \cdot -3x) \ &= \log(-15x) \end{align*}

So our equation now looks like this:

\begin{align*} \log(-15x) = 2 \end{align*}

Combining logarithms is a crucial step in simplifying logarithmic equations. By using properties like ${\log(a) + \log(b) = \log(ab)}$, we can reduce multiple logarithmic terms into a single, more manageable term. This not only simplifies the equation but also makes it easier to isolate the variable. In our specific case, combining ${\log(5)}$ and ${\log(-3x)}$ into ${\log(-15x)}$ allows us to move forward by eliminating one of the logarithmic terms. Remember, though, that when combining logarithms, you must ensure that the arguments of the logarithms are positive, as the logarithm of a non-positive number is undefined. This is something we'll need to verify later when we find a potential solution for ${x}$.

Step 2: Convert to Exponential Form

Now that we have a single logarithm, we need to convert the equation from logarithmic form to exponential form. Recall that ${\log_b(a) = c}$ is equivalent to ${b^c = a}$. In our equation, the base of the logarithm is 10 (since it's not explicitly written, it's a common logarithm). So we can rewrite the equation as:

\begin{align*} 10^2 = -15x \end{align*}

Which simplifies to:

\begin{align*} 100 = -15x \end{align*}

Converting to exponential form is a fundamental technique for solving logarithmic equations. By understanding the relationship between logarithms and exponents, we can transform a logarithmic equation into a more familiar algebraic equation. This conversion is based on the definition of a logarithm: if ${\log_b(a) = c}$, then ${b^c = a}$. In simpler terms, the logarithm ${c}$ is the exponent to which we must raise the base ${b}$ to obtain the value ${a}$. In our equation, ${\log(-15x) = 2}$ implies that ${10^2 = -15x}$, because the base of the common logarithm is 10. This step is essential because it allows us to eliminate the logarithm and work with a straightforward algebraic equation.

Step 3: Solve for x

Alright, now let's solve for ${x}$. Divide both sides of the equation by -15:

\begin{align*} \frac{100}{-15} = x \end{align*}

Simplify the fraction:

\begin{align*} x = -\frac{20}{3} \end{align*}

So our potential solution is ${x = -\frac{20}{3}}$.

Solving for x involves isolating the variable on one side of the equation. This typically involves performing algebraic operations such as addition, subtraction, multiplication, and division to both sides of the equation to maintain equality. In our case, after converting the logarithmic equation to exponential form, we obtained the equation ${100 = -15x}$. To isolate ${x}$, we divided both sides of the equation by -15, resulting in ${x = \frac{100}{-15}}$. Simplifying this fraction gives us ${x = -\frac{20}{3}}$. This is a critical step because it provides us with a potential value for ${x}$ that satisfies the original equation. However, it is important to remember that this is just a potential solution, and we must verify it to ensure it is valid.

Step 4: Check the Solution

We need to check if our solution is valid. Plug ${x = -\frac{20}{3}}$ back into the original equation:

\begin{align*} \log(5) + \log(-3(-\frac{20}{3})) = 2 \end{align*}

Simplify:

\begin{align*} \log(5) + \log(20) = 2 \end{align*}

Combine the logarithms again:

\begin{align*} \log(5 \cdot 20) = 2 \ \log(100) = 2 \end{align*}

Since ${\log(100) = 2}$, our solution is valid!

Checking the solution is a crucial step in solving logarithmic equations because not all potential solutions are valid. Logarithmic functions are only defined for positive arguments. Therefore, any value of ${x}$ that results in a non-positive argument for any logarithm in the original equation must be discarded. In our case, we found a potential solution of ${x = -\frac{20}{3}}$. To verify this solution, we substitute it back into the original equation: ${\log(5) + \log(-3x) = 2}$. This gives us ${\log(5) + \log(-3(-\frac{20}{3})) = \log(5) + \log(20)}$. Since both 5 and 20 are positive, the logarithms are defined. Combining the logarithms, we get ${\log(5 \cdot 20) = \log(100)}$. Since ${\log(100) = 2}$, our solution satisfies the original equation and is therefore valid. This step ensures that our final answer is mathematically correct and applicable to the original problem.

Step 5: Verify with a Graphing Calculator

To verify our solution graphically, we'll graph both sides of the original equation as two separate functions:

\begin{align*} y_1 = \log(5) + \log(-3x) \ y_2 = 2 \end{align*}

Using a graphing calculator, input these two equations. You should see that the graphs intersect at one point. Find the x-coordinate of that point of intersection. It should be approximately -6.67, which is the decimal representation of ${-\frac{20}{3}}$.

This confirms our solution!

Verifying with a graphing calculator provides a visual confirmation of our algebraic solution. By graphing both sides of the original equation as separate functions, we can identify the point of intersection. The x-coordinate of this point represents the value of ${x}$ for which both sides of the equation are equal, thus representing the solution. In our case, we graphed ${y_1 = \log(5) + \log(-3x)}$ and ${y_2 = 2}$. The intersection point should occur at approximately ${x = -6.67}$, which corresponds to our algebraic solution of ${x = -\frac{20}{3}}$. This graphical verification enhances our confidence in the correctness of our solution and demonstrates the relationship between algebraic and graphical methods in solving equations. It also helps in identifying any potential errors in our algebraic manipulations.

Final Answer

So, the solution to the equation ${\log(5) + \log(-3x) = 2}$ is:

\begin{align*} x = -\frac{20}{3} \end{align*}

And we've verified it both algebraically and graphically. Great job, guys! You nailed it!

Woo-hoo! Solving logarithmic equations can be a breeze once you get the hang of it. Remember to always check your solutions and use those graphing calculators to double-check your work. Keep up the awesome work, and I'll catch you in the next math adventure!