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Math Help - problems with logs and a proof.

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    problems with logs and a proof.

    1. I used the intermediate value theorem for (square root x to the third)= x^2+2x-1 on (0,1), I plugged 0 and 1, and I got (left side)0=-1(right side) and 1=2 respectively. Since I got a negative(on 0), and positive on the other(on 1), there is at least one solution for (0,1), is this right? If I have the wrong idea, tell me. 2. 1/2 lnx/ln3=ln8/ln2. ans. 729. 3. (square root to the third 2)^x + 10=2^(x^2). ans. 2, -5/3 4.(ln x/ln 3)+(ln 2x+1/ln 3)=1 ans. 1 I do not know how to put the small numbers on logs so I put it in natural log form. Thanks in advance.
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    Quote Originally Posted by driver327 View Post
    1. I used the intermediate value theorem for (square root x to the third)= x^2+2x-1 on (0,1), I plugged 0 and 1, and I got (left side)0=-1(right side) and 1=2 respectively. Since I got a negative(on 0), and positive on the other(on 1), there is at least one solution for (0,1), is this right? If I have the wrong idea, tell me.
    Wow. Ummmm... Let me break this down a bit. You've got the function:
    \sqrt{x^3} = x^2 + 2x - 1
    on the interval (0, 1).

    Are you trying to find a solution for x?

    If so then try moving everything to the left:
    \sqrt{x^3} - (x^2 + 2x - 1) = 0

    NOW plug in x = 0 and x = 1:
    \sqrt{0^3} - (0^2 + 2 \cdot 0 - 1) = 1

    \sqrt{1^3} - (1^2 + 2 \cdot 1 - 1) = -1

    Because the first is positive and the second negative we know that it (the LHS) passes through 0 at some point in the interval.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by driver327 View Post
    2. 1/2 lnx/ln3=ln8/ln2. ans. 729.
    Again, solve for x?

    \frac{1}{2} \frac{ln(x)}{ln(3)} = \frac{ln(8)}{ln(2)}

    Now,
    ln(8) = ln(2^3) = 3 \cdot ln(2), so

    \frac{1}{2} \frac{ln(x)}{ln(3)} = \frac{3 \cdot ln(2)}{ln(2)}

    \frac{1}{2} \frac{ln(x)}{ln(3)} = 3

    \frac{ln(x)}{ln(3)} = 6

    ln(x) = 6 \cdot ln(3)

    ln(x) = ln(3^6) = ln(729)

    x = 729

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by driver327 View Post
    3. (square root to the third 2)^x + 10=2^(x^2). ans. 2, -5/3
    Is this (\sqrt{2}~^3)^x + 10 = 2^{x^2}?

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by driver327 View Post
    4.(ln x/ln 3)+(ln 2x+1/ln 3)=1 ans. 1
    Is this supposed to be
    \frac{ln(x)}{ln(3)} + \frac{ln(2x + 1)}{ln(3)} = 1?

    ln(x) + ln(2x + 1) = ln(3)

    ln(x(2x + 1)) = ln(3)

    x(2x + 1) = 3

    2x^2 + x - 3 = 0

    (2x + 3)(x - 1) = 0

    So
    x = -\frac{3}{2}
    or
    x = 1

    But a negative x is outside the domain of the expression on the LHS. Thus the only solution is x = 1.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Is this (\sqrt{2}~^3)^x + 10 = 2^{x^2}?

    -Dan
    The 3 was on top of the square root, and the +10 was part of the exponent. Sorry for the confusion.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by driver327 View Post
    The 3 was on top of the square root, and the +10 was part of the exponent. Sorry for the confusion.
    So it's
    \sqrt[3]{2^{(x+10)}} = 2^{x^2} <-- This I can work with!

    Well, \sqrt[3]{x} = x^{1/3} so:
    (2^{(x+10)})^{1/3} = 2^{x^2}

    (2^{(x+10)/3}) = 2^{x^2}

    Thus
    \frac{x + 10}{3} = x^2

    x + 10 = 3x^2

    3x^2 - x - 10 = 0

    (3x + 5)(x - 2) = 0

    Thus
    x = -\frac{5}{3} or x = 2.

    Both solutions are in the domain of the original expression, so both are solutions.

    -Dan
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