Question

In(x+1)-In(x)=1 i know the answer is 1over e-1 but how?

Answer 1

I rearranges it: ln(x+1)=1+ln(x)
Substitute 1=ln(e) and use the properties of logarithms to get ln(x+1)=ln(ex)
So x+1=ex
Solve for x

Answer 2

I recommend this way , too

\[e^{\ln(x+1)-\ln(x)}=e^1\]

by taking both sides to the exponent

Answer 3

so by property of exponents,

\[{e^{\ln(x+1)} \over e^{\ln(x)} } =e\]

and by definition of logarithm,

\[{x+1 \over x} = e\]

Answer 4

so by cross multiplication

x+1 = ex

x-ex = -1

x(1-e) = -1

x =\[-1 \over 1-e\]

Answer 5

if you multiply -1/-1 to make the top = 1,

\[1 \over e-1\]

Answer 6

let me know if there is anything that you didn't get

Answer 7

Hey Yuki if you could please elaborate a little on this part - " \[e^{\ln(x+1)}\div e^{\ln(x)}=e\]
and by definition of logarithm,
x+1/x = e"

How do you end up on the second equation from the first?