Question

PLEASE HELP
1. ln x - ln(x-2)=5

Answer 1

Some rules of logarithms:

\(\color{#000000 }{ \displaystyle \ln(a)+\ln(b)=\ln\left(a \times b\right) }\)

\(\color{#000000 }{ \displaystyle \ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right) }\)

\(\color{#000000 }{ \displaystyle a=a\times \ln(e)=\ln(e^a)}\)

\(\color{#000000 }{ \displaystyle \ln(a)=b\quad \Longrightarrow \quad e^b=a }\)

Answer 2

i'm still a little confused...

Answer 3

If I had something like, (for example)

\(\color{#000000 }{ \displaystyle \ln(z)-\ln(z+5)=10 }\)

then, I would do this:

\(\color{#000000 }{ \displaystyle \ln\left(\frac{z}{z+5}\right)=10 }\)
(via the 2nd rule i posted)

\(\color{#000000 }{ \displaystyle e^{10}=\frac{z}{z+5} }\)
(via the last rule i posted)

Then, algebra:

(1) Multiply both sides times \(z+5\)

\(\color{#000000 }{ \displaystyle (z+5)e^{10}=z }\)

(2) expand the left side

\(\color{#000000 }{ \displaystyle e^{10}z+5e^{10} =z }\)

(3) subtract \(e^{10}z\) from both sides.

\(\color{#000000 }{ \displaystyle 5e^{10} =z-e^{10}z }\)

(4) factor the right side

\(\color{#000000 }{ \displaystyle 5e^{10} =z }\)

(5) divide by \(\color{#000000 }{ \displaystyle (1-e^{10}) }\) on both sides

\(\color{#000000 }{ \displaystyle \frac{5e^{10} }{1-e^{10}}=z }\)

Answer 4

Note that \(e\) is just a constant.

Answer 5

oh ok thanks for clarifying i was confused with e part

Answer 6

Also, very important property,
\(\color{#000000 }{ \displaystyle \log_a(a)=1 }\)

(this is true when "a" is any positive number, except 1)

Same way,
\(\color{#000000 }{ \displaystyle \ln(e)=\log_e(e)=1 }\)
(because \(\color{#000000 }{ \displaystyle \ln }\) is just a log with base of "e")

Answer 7

is e the base? in your explanation of how to solve the problem

Answer 8

yes, everytime you see ln, it is same as log\(_e\),

if that is what you are asking

Answer 9

yes it is :)

Answer 10

im doing the calculations now

Answer 11

i got 2.0135....

Answer 12

i checked it and it seems about right i think

Answer 13

mind helping with some more?

Answer 14

Yes, if I have time, but I need to ask you a question thougfh...

Answer 15

yes?

Answer 16

don't you need to find the exact solution for x?
(not just an aproximation)

Answer 17

no my teacher doesn't mind. i just put a couple numbers after the decimal point and she doesn't mind

Answer 18

Ok:)

Answer 19

i posted another question if you have time :)