Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

7. Find a number t such that ln(2t + 1) = −4 exact value?

precalc

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  • 2 months ago

    ln(2t+1)=-4

    =>

    2t+1=e^(-4)

    =>

    2t=e^(-4)-1

    =>

    2t=(1-e^4)/e^4

    =>

    t=(1-e^4)/(2e^4)

  • 2 months ago

    ln(2t + 1) = -4

    Making each side an exponent over a base of "e" will cancel out the natural log and leave only what's inside.  Doing that we get:

    2t + 1 = e⁻⁴

    Now solving for t:

    2t = e⁻⁴ - 1

    t = (e⁻⁴ - 1) / 2

    or:

    t = 1/(2e⁴) - 1/2

    That's your exact value.  A decimal approximation to that would be:

    t = -0.4908422

    Substituting this into the original equation we can test this value:

    ln(2t + 1) = -4

    ln[2(-0.4908422) + 1] = -4

    ln(-0.9816844 + 1) = -4

    ln(0.0183156) = -4

    -4.000002 = -4

    Since we rounded to 7SF, we should be correct to 6SF.  Checking the first 6 SF of the left side it does equal the right side.  So this is correct.

    Again the answer is:

    t = 1/(2e⁴) - 1/2

  • fcas80
    Lv 7
    2 months ago

    t can be solved in terms of e.  e is irrational.  The resulting expression will be exact in terms of e, but it will not be an integer.

  • 2 months ago

    ln(2t + 1) = -4

    so, 2t + 1 = e⁻⁴

    Hence, 2t = e⁻⁴ - 1

    so, t = (e⁻⁴ - 1)/2

    :)>

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  • rotchm
    Lv 7
    2 months ago

    Hint: do e^ each side. What does that give?

    Un-anon yourself & we will detail further.

    Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. 

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