Math  /  Algebra

QuestionSolve the following logarithmic equation. Express your answer as either an exact expression or a decimal approximation rounded to four decimal places. If there is no solution, indicate "No Solution ( \emptyset )." ln(x+6)+ln(x)=0\ln (x+6)+\ln (x)=0

Studdy Solution

STEP 1

What is this asking? We need to find the value(s) of xx that make the equation ln(x+6)+ln(x)=0\ln(x+6) + \ln(x) = 0 true. Watch out! Remember that the natural logarithm, ln\ln, is only defined for *positive* inputs.
So, any solution we find must make both xx and x+6x+6 greater than zero!

STEP 2

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

STEP 3

We can combine the two logarithms on the left side of the equation using the product rule: ln(a)+ln(b)=ln(ab)\ln(a) + \ln(b) = \ln(a \cdot b).
So, ln(x+6)+ln(x)\ln(x+6) + \ln(x) becomes ln((x+6)x)\ln((x+6) \cdot x), which simplifies to ln(x2+6x)\ln(x^2 + 6x).
Our equation is now ln(x2+6x)=0\ln(x^2 + 6x) = 0.
This is much nicer to work with!

STEP 4

Remember that ln\ln means the logarithm with base ee.
To get rid of the logarithm, we can rewrite the equation in exponential form.
If ln(a)=b\ln(a) = b, then eb=ae^b = a.
In our case, ln(x2+6x)=0\ln(x^2 + 6x) = 0, so e0=x2+6xe^0 = x^2 + 6x.

STEP 5

Since *anything* to the power of **zero** is **one** (except 000^0, which is undefined), we have e0=1e^0 = 1.
So our equation becomes 1=x2+6x1 = x^2 + 6x.

STEP 6

Let's rearrange our equation into standard quadratic form: ax2+bx+c=0ax^2 + bx + c = 0.
Subtracting **one** from both sides of 1=x2+6x1 = x^2 + 6x gives us 0=x2+6x10 = x^2 + 6x - 1, or x2+6x1=0x^2 + 6x - 1 = 0.

STEP 7

We can solve for xx using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
In our equation, a=1a=1, b=6b=6, and c=1c=-1.
Plugging these values into the quadratic formula, we get: x=6±6241(1)21x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} x=6±36+42x = \frac{-6 \pm \sqrt{36 + 4}}{2}x=6±402x = \frac{-6 \pm \sqrt{40}}{2}x=6±2102x = \frac{-6 \pm 2\sqrt{10}}{2}x=3±10x = -3 \pm \sqrt{10}So, our potential solutions are x=3+10x = -3 + \sqrt{10} and x=310x = -3 - \sqrt{10}.

STEP 8

Remember, we can only take the logarithm of *positive* numbers!
We need to check if our potential solutions make both xx and x+6x + 6 positive.

STEP 9

For x=3+103+3.16=0.16x = -3 + \sqrt{10} \approx -3 + 3.16 = 0.16, we have x>0x > 0 and x+60.16+6=6.16>0x + 6 \approx 0.16 + 6 = 6.16 > 0.
So, x=3+10x = -3 + \sqrt{10} is a valid solution!

STEP 10

For x=31033.16=6.16x = -3 - \sqrt{10} \approx -3 - 3.16 = -6.16, we have x<0x < 0.
Since xx must be positive, this solution is not valid.

STEP 11

The solution to the equation ln(x+6)+ln(x)=0\ln(x+6) + \ln(x) = 0 is x=3+10x = -3 + \sqrt{10}.

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