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Math Help - logs and exponents of natural base

  1. #1
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    logs and exponents of natural base

    I'm having trouble trying to solve for x in these equations, and I have a calculus exam on monday, so any help would be great!!

    1. Solve each equation
    (a) e^x - 2e^-x = 1

    (b) ln(x^2 + e) - ln (x + 1) = 1



    If you could explain the steps in getting the answers, that would be much appreciated. I'm a little lost on how to start them off.

    The answers we were given on the sheet are:
    (a) ln2 (b) 0, e
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  2. #2
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    Quote Originally Posted by jmailloux View Post
    I'm having trouble trying to solve for x in these equations, and I have a calculus exam on monday, so any help would be great!!

    1. Solve each equation
    (a) e^x - 2e^-x = 1

    (b) ln(x^2 + e) - ln (x + 1) = 1



    If you could explain the steps in getting the answers, that would be much appreciated. I'm a little lost on how to start them off.

    The answers we were given on the sheet are:
    (a) ln2 (b) 0, e
    a) e^x -2e^(-x) = 1
    Rewriting that in fraction form,
    e^x -2/(e^x) = 1
    Clear the fraction, multiply both sides by e^x,
    (e^x)^2 -2 = e^x
    (e^x)^2 -e^x -2 = 0
    (e^x -2)(e^x +1) = 0
    e^x = 2 or -1

    When e^x = 2,
    x*ln(e) = ln(2)
    x = ln(2) ---------------------**

    When e^x = -1
    x*ln(e) = ln(-1)
    Uh-oh, there is no log of negative numbers, so disregard e^x = -1

    Therefore, x = ln(2) ------------answer.

    ---------------------------------------------------
    b) ln(x^2 +e) -ln(x +1) = 1
    ln[(x^2 +e) /(x +1)] = ln(e)
    (x^2 +e) /(x +1) = e
    x^2 +e = e(x+1)
    x^2 +e = ex +e
    x^2 = ex
    x^2 -ex = 0
    x(x -e) = 0
    x = 0 or e. ----------------answer.
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  3. #3
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    Quote Originally Posted by jmailloux View Post
    I'm having trouble trying to solve for x in these equations, and I have a calculus exam on monday, so any help would be great!!

    1. Solve each equation
    (a) e^x - 2e^-x = 1
    Multiply through by e^x to get:

    (e^x)^2-2=e^x

    which is a quadratic in y=e^x which making this substitution and re arranging gives:

    y^2-y-2=0

    which can be solved using the quadratic formula or by noticing that it factors to give:

    y^2-y-2=(y+1)(y-2)=0

    which has roots y=-1,2, s the solutions of the original equation are x=\ln(-1) which is imposible (in the reals) and x=\ln(2).

    So the only solution in the real numbers is x=\ln(2).

    (b) ln(x^2 + e) - ln (x + 1) = 1
    Here we use the law of logarithms to combine the terms on the left:

    \ln(x^2 + e) - \ln (x + 1)=\ln \left( \frac{x^2+e}{x+1} \right) = 1

    Now we use the definition of natural logarithms to get rid of the logs:

    \frac{x^2+e}{x+1}=e^1=e

    which is a quadratic in x:

    x^2+e=e\,x+e

    rearranging:

    x^2-e\,x=0,

    so x=0, e.

    RonL
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  4. #4
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    thanks

    thanks.. it makes A LOT more sense now
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