Question

How do I solve ln x + ln x=0? Was never taught natural log or log definitions ever sorry;( Please help!!

Answer 1

ln x + ln x = 0 ?

Answer 2

Yes

Answer 3

2 ln(x) = 0

Answer 4

you can combine ln(x) + ln(x)
using
ln(a*b)= ln(a)+ln(b)
as a first step

Answer 5

then make each side the exponent to the base e

Answer 6

Okay but once I make each side the base of e what does e represent? Or how do I solve for e because I was never taught that either:( sorry probably sound really stupid to you guys right now.

Answer 7

you need to know
ln (natural log) is the same thing as log base e (e=2.718.....)

there is this rule
\[ e^{\ln(a)} = a\]

for your problem
ln(x)+ln(x) = 0
ln(x*x) = 0 (using the rule posted above)
ln(x^2) = 0 (another way to write x*x)

now make each side the exponent of base e
\[ e^{\ln(x^2)} = e^0 \]

using the rule \[ e^{\ln(a)} = a\]
this is the same as
\[ x^2 = e^0\]

any number to the zero power is 1 (important fact)
\[ x^2= 1 \\ x= \sqrt{1}= 1\]

you only use the positive root, because we do not take logs of negative numbers

Answer 8

see http://www.khanacademy.org/math/algebra/logarithms-tutorial/logarithm_basics/v/logarithms

for video(s) on logs... they might help.

Answer 9

Thank you SO much!! I really appreciate it

Answer 10

Here is Khan's whole list of videos on logs
http://www.khanacademy.org/math/algebra/logarithms-tutorial

It would take a big bag of popcorn to get through all of them, but watching a few might be good

Answer 11

Here is wolfram's solution
http://www.wolframalpha.com/input/?i=ln%28x%29%2Bln%28x%29%3D+0

Answer 12

thank you so much I really appreciate it because I had no idea