Question $\log_7(x-1) + \log_7(x+3) = \log_7(x+2)$

Studdy Solution

STEP 1

What is this asking? We need to find the value of $x$ that makes this logarithmic equation true! Watch out! Remember that the arguments of logarithms must be positive, so we'll need to check our solutions to make sure they don't break this rule.

STEP 2

1. Combine the logs
2. Rewrite as an exponential equation
3. Solve for $x$
4. Check our solution

STEP 3

We've got a sum of logs with the same base, so we can use the product rule: $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$ .
This lets us combine the left side of the equation: $\log_7((x-1)(x+3)) = \log_7(x+2)$ This is awesome because it simplifies things a lot!

STEP 4

Let's expand the product inside the logarithm on the left side: $\log_7(x^2 + 2x - 3) = \log_7(x+2)$ Now we have a single log on each side, which is much easier to work with.

STEP 5

Since both sides of the equation have a $\log$ with base 7, we can set the arguments equal to each other: $x^2 + 2x - 3 = x + 2$ This gets rid of the logs and gives us a regular quadratic equation to solve.

STEP 6

Let's move all the terms to one side to set the equation to zero: $x^2 + x - 5 = 0$ Now we have a quadratic equation in standard form.

STEP 7

We can use the quadratic formula to solve for $x$ : $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ In our equation, $a = \mathbf{1}$ , $b = \mathbf{1}$ , and $c = \mathbf{-5}$ .
Plugging these values into the quadratic formula, we get: $x = \frac{-\mathbf{1} \pm \sqrt{\mathbf{1}^2 - 4 \cdot \mathbf{1} \cdot (\mathbf{-5})}}{2 \cdot \mathbf{1}}$ $x = \frac{-1 \pm \sqrt{1 + 20}}{2}$ $x = \frac{-1 \pm \sqrt{21}}{2}$ So, we have two possible solutions: $x = \frac{-1 + \sqrt{21}}{2}$ and $x = \frac{-1 - \sqrt{21}}{2}$ .

STEP 8

Remember, the arguments of our logarithms must be positive!
Let's check both solutions.

STEP 9

For $x = \frac{-1 + \sqrt{21}}{2} \approx 1.79$ , we have: $x - 1 \approx 0.79 > 0$ $x + 3 \approx 4.79 > 0$ $x + 2 \approx 3.79 > 0$ Since all arguments are positive, this solution works!

STEP 10

For $x = \frac{-1 - \sqrt{21}}{2} \approx -2.79$ , we have: $x - 1 \approx -3.79 < 0$ Oops! Since $x-1$ is negative, this solution is not valid.

STEP 11

The solution to the equation $\log_7(x-1) + \log_7(x+3) = \log_7(x+2)$ is $x = \frac{-1 + \sqrt{21}}{2}$ .

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