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FreeTrader

  • 3 years ago

Can you please help guide me through the last steps in this equation involving logarithms?

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  1. FreeTrader
    • 3 years ago
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    \[1+\ln x = \ln4\]

  2. FreeTrader
    • 3 years ago
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    I am solving for x. Can you nudge me about the ln properties?

  3. jim_thompson5910
    • 3 years ago
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    1+ln(x) = ln(4) 1 = ln(4)-ln(x) 1 = ln(4/x) 4/x = e^1 4/x = e 4 = ex 4/e = x x = 4/e

  4. FreeTrader
    • 3 years ago
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    Thanks, I'll see if I can follow this on paper here... Just a sec...

  5. jim_thompson5910
    • 3 years ago
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    The properties I used were ln(x) - ln(y) = ln(x/y) and ln(x) = y converts to e^y = x

  6. FreeTrader
    • 3 years ago
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    I see the division coming into play with the subtraction.

  7. FreeTrader
    • 3 years ago
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    The last one is the key I was missing.

  8. satellite73
    • 3 years ago
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    you can also try \[\ln(x)=\ln(4)-1\] \[x=e^{\ln(4)-1}=e^{\ln(4)}\times e^{-1}=\frac{4}{e}\]

  9. FreeTrader
    • 3 years ago
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    ln x = e to what power gives us x Logarithms confuse me to no end, despite how "simple" they are supposed to be.

  10. FreeTrader
    • 3 years ago
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    I have duplicated the algebra, and it makes sense. Are you aware of a really good ln tutorial with exercises?

  11. jim_thompson5910
    • 3 years ago
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    try this page: http://www.purplemath.com/modules/solvelog.htm

  12. FreeTrader
    • 3 years ago
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    Thank you. If you answer that tutorial question in the new Q I posted, I'd be pleased to pin a medal upon you.

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