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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    I recommend this way , too \[e^{\ln(x+1)-\ln(x)}=e^1\] by taking both sides to the exponent

  3. Yuki
    • 5 years ago
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    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
    • 5 years ago
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    so by cross multiplication x+1 = ex x-ex = -1 x(1-e) = -1 x =\[-1 \over 1-e\]

  5. Yuki
    • 5 years ago
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    if you multiply -1/-1 to make the top = 1, \[1 \over e-1\]

  6. Yuki
    • 5 years ago
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    let me know if there is anything that you didn't get

  7. anonymous
    • 5 years ago
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    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?

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