Question 251163: How would I solve the a natural log problem like the one below?
Thank you!!!
((lnX)^2)-6 = -5lnX
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! problem is:
(ln(x))^2 - 6 = -5*ln(x)
add 5*ln(x) to both sides of this equation to get:
(ln(x))^2 + 5*ln(x) - 6 = 0
let y = ln(x) to get:
y^2 + 5y - 6 = 0
factor this equation to get:
(y+6)*(y-1) = 0
solve for y to get:
y = -6 or y = 1
substitute ln(x) for y to get
ln(x) = -6 or ln(x) = 1
by the laws of logarithms, ln(x) = y if and only if e^y = x
based on that law, we get:
ln(x) = -6 if and only if e^-6 = x
and we get:
ln(x) = 1 if and only if e^1 = x
we will solve for each separately.
first we'll solve for e^-6 = x
using your calculator, e^-6 = .002478752 which makes:
x = .002478752.
next we'll solve for e^1 = x
using your calculator, e^1 = 2.718281828 which makes:
x = 2.718281828
x = .002478752 or x = 2.718281828
we can confirm by substituting in the original equations.
it is:
(ln(x))^2 - 6 = -5*ln(x)
for x = 2.718281828, this equation becomes:
(ln(2.718281828))^2 - 6 = -5*ln(2.718281828)
since ln(2.718281828) = 1, this equation becomes:
1^2 - 6 = -5 which is true, confirming that x = 2.718281828 is good.
for x = .002478752, this equation becomes:
(ln(.002478752))^2 - 6 = -5*ln(.002478752)
since ln(.002478752 = -6, this equation becomes:
(-6)^2 - 6 = -5*-6 which becomes:
36 - 6 = 30 which is true, confirming that x = .002478752 is good.
both answers are good.
your answer is:
x = 2.718281828
or:
x = .002478752
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