anonymous 5 years ago In(x+1)-In(x)=1 i know the answer is 1over e-1 but how?

1. anonymous

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

2. Yuki

I recommend this way , too $e^{\ln(x+1)-\ln(x)}=e^1$ by taking both sides to the exponent

3. Yuki

so by property of exponents, ${e^{\ln(x+1)} \over e^{\ln(x)} } =e$ and by definition of logarithm, ${x+1 \over x} = e$

4. Yuki

so by cross multiplication x+1 = ex x-ex = -1 x(1-e) = -1 x =$-1 \over 1-e$

5. Yuki

if you multiply -1/-1 to make the top = 1, $1 \over e-1$

6. Yuki

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

7. anonymous

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?