Sure, let's solve the logarithmic equation:
[tex]\[
\ln x + \ln (x + 8) = 1
\][/tex]
### Step-by-Step Solution:
1. Combine the logarithms using the property of logarithms:
The sum of logarithms can be combined into a single logarithm:
[tex]\[
\ln x + \ln (x + 8) = \ln [x(x + 8)]
\][/tex]
So the equation becomes:
[tex]\[
\ln [x(x + 8)] = 1
\][/tex]
2. Rewrite the equation in exponential form:
Recall that if [tex]\(\ln a = b\)[/tex], then [tex]\(e^b = a\)[/tex]. Here, [tex]\(a = x(x + 8)\)[/tex] and [tex]\(b = 1\)[/tex]. Thus:
[tex]\[
x(x + 8) = e^1
\][/tex]
Since [tex]\(e^1\)[/tex] is simply [tex]\(e\)[/tex], the equation becomes:
[tex]\[
x(x + 8) = e
\][/tex]
3. Simplify and solve the quadratic equation:
Rearrange the equation into standard quadratic form:
[tex]\[
x^2 + 8x - e = 0
\][/tex]
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = -e\)[/tex].
4. Use the quadratic formula to find [tex]\(x\)[/tex]:
The quadratic formula is:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
Substituting [tex]\(a = 1\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = -e\)[/tex]:
[tex]\[
x = \frac{{-8 \pm \sqrt{{8^2 - 4(1)(-e)}}}}{2(1)}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{{-8 \pm \sqrt{{64 + 4e}}}}{2}
\][/tex]
Further simplify:
[tex]\[
x = \frac{{-8 \pm \sqrt{{64 + 4e}}}}{2}
\][/tex]
[tex]\[
x = -4 \pm \frac{{\sqrt{{64 + 4e}}}}{2}
\][/tex]
[tex]\[
x = -4 \pm \sqrt{{16 + e}}
\][/tex]
5. Evaluate the possible solutions:
Since we need the solution to be positive (as the logarithm [tex]\(\ln x\)[/tex] is defined only for positive [tex]\(x\)[/tex]), let’s consider:
[tex]\[
x = -4 + \sqrt{{16 + e}}
\][/tex]
Evaluate this expression (using the known value of [tex]\(e\)[/tex] approximately 2.718):
[tex]\[
x \approx -4 + \sqrt{{16 + 2.718}}
\][/tex]
[tex]\[
x \approx -4 + \sqrt{{18.718}}
\][/tex]
Evaluate the square root:
[tex]\[
x \approx -4 + 4.326
\][/tex]
Finally:
[tex]\[
x \approx 0.326
\][/tex]
Thus, the solution to the logarithmic equation [tex]\(\ln x + \ln (x + 8) = 1\)[/tex] is:
[tex]\[
x \approx 0.326
\][/tex]
