QuestionSolve the logarithmic equation. $\log _{9}(x-2)+\log _{9} x=\frac{1}{2}$

Studdy Solution

STEP 1

What is this asking? We're trying to find a mystery number $x$ that makes this logarithmic equation true! Watch out! Remember logarithmic properties and watch out for values of $x$ that would make the arguments of the logarithms zero or negative!

STEP 2

1. Combine the logarithms
2. Rewrite in exponential form
3. Solve the quadratic equation
4. Check for extraneous solutions

STEP 3

We can combine the two logarithms on the left side of the equation using the product property of logarithms, which states that $\log_b{m} + \log_b{n} = \log_b{(m \cdot n)}$ .
In our case, $b = 9$ , $m = (x - 2)$ , and $n = x$ .
So, we get: $\log_9{((x-2) \cdot x)} = \frac{1}{2}$ $\log_9{(x^2 - 2x)} = \frac{1}{2}$ Why did we do this?
Combining the logarithms makes the equation simpler and gets us closer to isolating $x$ !

STEP 4

We can rewrite the logarithmic equation in exponential form.
Remember, $\log_b{a} = c$ is equivalent to $b^c = a$ .
In our case, $b = 9$ , $a = x^2 - 2x$ , and $c = \frac{1}{2}$ .
So, we have: $9^{\frac{1}{2}} = x^2 - 2x$ This step is crucial because it transforms the equation into a more familiar form that we can solve more easily!

STEP 5

Remember that $9^{\frac{1}{2}}$ is the same as $\sqrt{9}$ , which equals 3.
So, our equation becomes: $3 = x^2 - 2x$ See how much simpler our equation looks now?

STEP 6

Let's rearrange the equation to get a standard quadratic equation in the form $ax^2 + bx + c = 0$ .
We can subtract 3 from both sides of the equation: $x^2 - 2x - 3 = 0$ This sets the stage for solving for $x$ !

STEP 7

We can factor the quadratic equation as follows: $(x - 3)(x + 1) = 0$ Factoring helps us find the values of $x$ that satisfy the equation.

STEP 8

Now, we set each factor equal to zero and solve for $x$ : $x - 3 = 0 \implies x = 3$ $x + 1 = 0 \implies x = -1$ So, our potential solutions are $x = 3$ and $x = -1$ .

STEP 9

Remember, we can't have a negative or zero value inside a logarithm.
Let's check our potential solutions in the original equation: * For $x = 3$ : $\log_9(3-2) + \log_9(3) = \log_9(1) + \log_9(3) = 0 + \frac{1}{2}\log_3(3) = \frac{1}{2}$ .
This works! * For $x = -1$ : $\log_9(-1-2) + \log_9(-1)$ .
Uh oh!
We have negative numbers inside the logarithms, which isn't allowed.
So, $x = -1$ is an extraneous solution.

STEP 10

Therefore, the only valid solution is $x = 3$ !

STEP 11

The solution to the logarithmic equation $\log _{9}(x-2)+\log _{9} x=\frac{1}{2}$ is $x = 3$ .

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