Shadow:

Logarithim Problem

8 months ago
Shadow:

Gimme a sec

8 months ago
Shadow:

\[\ln (x + 2) - \ln x = \ln (x + 5)\]

8 months ago
Shadow:

Just asking to solve.

8 months ago
Vocaloid:

|dw:1508207631214:dw|

8 months ago
Vocaloid:

log rule 2, combine the ln(x+2) and the ln(x) on the left into one ln

8 months ago
Shadow:

\[\ln \frac{ x + 2 }{ x } = \ln (x + 5)\]

8 months ago
Vocaloid:

good, now when you raise both sides to the power of e you just get (x+2)/x = x + 5 then it's just algebra to solve for x

8 months ago
Vocaloid:

let me know what you get

8 months ago
Shadow:

Sorry. Got pulled away. \[\frac{ x + 2 }{ x } = x + 5\] \[x + 2 = x^2 + 5x\] \[x^2 + 4x - 2 = 0\] Quadratic Formula \[\frac{ -4 \pm \sqrt (4^2 -4(1)(-2) }{ 2(1) }\] \[\frac{ -4 \pm \sqrt24 }{ 2 } \] \[\frac{ -4 \pm 2 \sqrt 6 }{ 2 }\] \[-2 \pm \sqrt 6\]

8 months ago
Vocaloid:

good, that's what I end up getting too

8 months ago
Vocaloid:

I believe you also have to check for extraneous solutions though

8 months ago
Vocaloid:

so you would plug in -2 + root(6) and -2 - root(6) into both equations to make sure both ln terms are defined

8 months ago
Shadow:

Oh yeah

8 months ago
Shadow:

Only the positive one works for me.

8 months ago
Vocaloid:

yup that's what I got too, gj

8 months ago
Shadow:

If I may ask, how did you check for extraneous solutions.

8 months ago
Vocaloid:

we have an ln(x) term so x must be positive

8 months ago
Vocaloid:

that automatically eliminates the -2 - root(6) solution then I just plugged in -2 + root(6) into the original equation and the math worked out fine

8 months ago
Shadow:

The polynomial or the first ln equation

8 months ago
Vocaloid:

both should work

8 months ago
Vocaloid:

ln(-2 + sqrt(6) + 2) - ln(-2+sqrt(6)) = ln(-2 + sqrt(6) + 5)) if you are using a calculator be careful with parentheses

8 months ago
Vocaloid:

both sides are ~1.695521979

8 months ago
Shadow:

Yeah, I was getting something returned weird, and alas, it was a parentheses. Thank you :)

8 months ago