Logarithm Equation Solved: Log3(x) + Log3(x-26) = 3

Hey math whizzes! Ever feel like tackling a logarithm equation is like trying to decipher an ancient scroll? Well, fear not, because today we're diving deep into the world of logs to conquer the equation $\log _3(x)+\log _3(x-26)=3$. This isn't just about finding 'x'; it's about understanding the why behind each step. We'll break it down, solve it, and then, the crucial part, check our answer to make sure we haven't gone down a rabbit hole of errors. So grab your calculators, your notebooks, and let's get this log party started!

Understanding the Logarithmic Foundation

Before we jump into solving, let's get our heads around what we're dealing with. A logarithm, in simple terms, is the inverse operation to exponentiation. Think of it like this: if $2^3 = 8$, then $\log _2(8) = 3$. The logarithm asks, "To what power do I need to raise the base (in this case, 2) to get the number (in this case, 8)?" The answer is the exponent (3).

In our equation, we have $\log _3(x)$ and $\log _3(x-26)$. The base is 3. This means we're looking for the power we need to raise 3 to in order to get 'x' and 'x-26', respectively. But there's a catch with logarithms, guys: the argument of a logarithm (the part inside the parentheses) must always be positive. This is super important and will come back to bite us later if we forget.

So, from $\log _3(x)$, we know that $x > 0$. And from $\log _3(x-26)$, we know that $x-26 > 0$, which simplifies to $x > 26$. For both of these conditions to be true, 'x' must be greater than 26. Keep this 'domain restriction' in mind – it's our first line of defense against extraneous solutions.

Unlocking the Equation: Logarithm Properties to the Rescue!

Now, let's look at the equation itself: $\log _3(x)+\log _3(x-26)=3$. See those two log terms added together? This is where our awesome logarithm properties come into play. The key property we need here is the product rule for logarithms: $\log _b(M) + \log _b(N) = \log _b(M \times N)$. Basically, when you add logs with the same base, you can combine them into a single log by multiplying their arguments.

Applying this to our equation, we get: $\log _3(x \times (x-26)) = 3$. This simplifies the left side of the equation, making it much more manageable. We've turned a sum of two logs into a single log. Pretty neat, huh?

Converting to Exponential Form

We're now at $\log _3(x(x-26)) = 3$. Our goal is to isolate 'x', and right now, 'x' is trapped inside a logarithm. To free it, we need to convert this logarithmic equation into its equivalent exponential form. Remember our definition of a logarithm? If $\log _b(A) = C$, then $b^C = A$.

In our case, the base '$b$' is 3, the argument '$A$' is $x(x-26)$, and the result '$C$' is 3. So, applying the conversion, we get: $3^3 = x(x-26)$.

Calculate $3^3$: $3 \times 3 \times 3 = 27$. So, our equation becomes $27 = x(x-26)$.

Solving the Quadratic Equation

We're not done yet, guys! We have $27 = x(x-26)$. Let's distribute the 'x' on the right side: $27 = x^2 - 26x$. To solve this, we need to rearrange it into the standard quadratic form, $ax^2 + bx + c = 0$. We can do this by subtracting 27 from both sides:

$0 = x^2 - 26x - 27$

Or, written the other way around: $x^2 - 26x - 27 = 0$.

Now we have a quadratic equation! There are a few ways to solve quadratics: factoring, using the quadratic formula, or completing the square. Factoring is often the quickest if it works. We need two numbers that multiply to -27 and add up to -26.

Let's think... The numbers are -27 and +1. Why? Because $(-27) \times (+1) = -27$ and $(-27) + (+1) = -26$. Perfect!

So, we can factor the quadratic equation as: $(x - 27)(x + 1) = 0$.

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for 'x':

1. $x - 27 = 0 \implies x = 27$
2. $x + 1 = 0 \implies x = -1$

We've found two potential solutions: $x=27$ and $x=-1$. But hold on! Remember that crucial step we talked about earlier – the domain restriction?

The Critical Check: Eliminating Extraneous Solutions

This is where the checking part of "Solve and Check" becomes absolutely vital. We determined at the beginning that for our original equation $\log _3(x)+\log _3(x-26)=3$ to be defined, 'x' must be greater than 26.

Let's test our potential solutions:

• Test $x = 27$:

  • Is $x > 0$? Yes, $27 > 0$.
  • Is $x > 26$? Yes, $27 > 26$.
  • Since both conditions are met, $x=27$ is a valid solution.

• Test $x = -1$:

  • Is $x > 0$? No, $-1$ is not greater than $0$.
  • Is $x > 26$? No, $-1$ is not greater than $26$.
  • Since these conditions are not met, the argument of the logarithms would be negative or zero, which is undefined in the real number system. Therefore, $x=-1$ is an extraneous solution and must be discarded.

So, after all that work, our only legitimate solution is $x=27$. It's a common trap with logarithmic and radical equations – you might get solutions algebraically, but they don't actually work in the original problem. Always, always check your answers!

Final Verification: Plugging Back Into the Original Equation

To be completely sure, let's plug our valid solution, $x=27$, back into the original equation: $\log _3(x)+\log _3(x-26)=3$.

Substitute $x=27$:

$\log _3(27) + \log _3(27-26) = 3$

$\log _3(27) + \log _3(1) = 3$

Now, let's evaluate each log term:

• $\log _3(27)$: What power do we raise 3 to get 27? $3^1=3$, $3^2=9$, $3^3=27$. So, $\log _3(27) = 3$.
• $\log _3(1)$: What power do we raise 3 to get 1? Any non-zero number raised to the power of 0 is 1. So, $\log _3(1) = 0$.

Substitute these values back into the equation:

$3 + 0 = 3$

$3 = 3$

Boom! The equation holds true. This final verification confirms that $x=27$ is indeed the correct solution. It feels pretty awesome when everything lines up perfectly, right?

Key Takeaways for Your Math Arsenal

So, what have we learned from this logarithmic adventure, guys? Several key things:

1. Domain Restrictions are King: Always identify the values of 'x' that make the arguments of your logarithms positive before you start solving. This is your superpower against extraneous solutions.
2. Master Logarithm Properties: The product rule ($\log M + \log N = \log(MN)$) was essential here. Remember the quotient rule ($\log M - \log N = \log(M/N)$) and the power rule ($k \log M = \log(M^k)$) too – they'll unlock many other problems.
3. Convert to Exponential Form: When you have a single log term equal to a number, converting to exponential form ($b^C = A$) is your ticket to solving for 'x'.
4. Solve the Resulting Equation: Often, this means solving a quadratic equation. Be prepared to factor, use the quadratic formula, or complete the square.
5. CHECK YOUR ANSWERS: I cannot stress this enough! Always plug your potential solutions back into the original equation to ensure they are valid and not extraneous. This step is non-negotiable.

Solving equations like $\log _3(x)+\log _3(x-26)=3$ might seem intimidating at first glance, but by breaking it down into smaller, manageable steps and remembering the fundamental rules of logarithms and algebra, you can conquer it. Keep practicing, stay curious, and you'll become a logarithm master in no time! Happy solving!