SOLUTION: I am trying to figure out this question. ln(x+1)= ln(3x+1)-ln x??? I keep getting x= square root 2x+1. but I don't think that is write??? I am horrible at these. Please help!!!
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Question 79558This question is from textbook College Algebra Graphs and Models
: I am trying to figure out this question. ln(x+1)= ln(3x+1)-ln x??? I keep getting x= square root 2x+1. but I don't think that is write??? I am horrible at these. Please help!!! This question is from textbook College Algebra Graphs and Models
You can put this solution on YOUR website! ln(x+1)= ln(3x+1)-ln x
ln(x+1) = ln[(3x+1)/x]
Since the ln's are equal, the anti-ln's are equal:
x+1 = (3x+1)/x
x^2+x = 3x+1
x^2-2x-1=0
x=[2+-sqrt4-4*-1]/2
x=[2+-sqrt(8)]/2
x=[1+-sqrt2]
Positive answer:
x=1+sqrt2
x=2.1414....
===============
Cheers,
Stan H.
ln(x+1) = ln(3x+1) - ln x
ln(x+1) + ln(x) = ln(3x+1)
ln[(x+1)x] = ln(3x+1)
ln(x²+x) = ln(3x+1)
Use the rule: if ln(A) = ln(B) then A = B
x²+x = 3x+1
x² - 2x - 1 = 0
Get 0 on the right by subtracting 3x and 1
from both sides:
x² - 2x - 1 = 0
Use the quadratic formula:
______
-b ± Öb²-4ac
x = —————————————
2a
where a = 1; b = -2; c = -1
______________
-(-2) ± Ö(-2)²-4(1)(-1)
x = ————————————————————————
2(1)
___
2 ± Ö4+4
x = ———————————
2
_
2 ± Ö8
x = ————————
2
___
2 ± Ö4·2
x = ———————————
2
_
2 ± 2Ö2
x = ——————————
2
_
2 2Ö2
x = ——— ± —————
2 2
_
x = 1 ± Ö2
_
Using the +, x = 1 + Ö2, which
is one answer and equals about 2.141213562
_
Using the -, x = 1 - Ö2, which
is the other answer and equals about -.4142135624.
However since the original problem contains ln(x),
and since logarithms can only be taken of positive
numbers, the only solution is:
_
x = 1 + Ö2
Edwin