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Math Help - Is this correct?

  1. #1
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    Is this correct?

    The directions are...
    "Use the Change of Base Formula to solve the equation to the nearest thousanth:

    Is this correct?-math-problem.png
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  2. #2
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    Re: Is this correct?

    Quote Originally Posted by Mollers View Post
    The directions are...
    "Use the Change of Base Formula to solve the equation to the nearest thousanth:

    Click image for larger version. 

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    This is incorrect, you should have done this:

    \displaystyle \begin{align*} 11^x - 50 &= 12 \\ 11^x &= 62 \end{align*}

    Now take the natural logarithm of both sides.
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  3. #3
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    Re: Is this correct?

    thank you I got it!

    I feel dumb asking some of these because they look so basic but I want to make sure I know what I'm doing.

    1. The domain and range of 5x cubed + 4
    I got negative infinite - positive infinite.

    2. Write as exponential form: ln 1 = 0
    I got 10^0 = 1

    3. Use properties of logarithms to rewrite the expression: 4 log base 2 a squared
    I got log base 2 a^8

    4. Solve the equation: log x = -2
    I got 10^-2 = .01
    X = .01

    5. 2 ln y = 9
    I got 2y = ln 9
    2y = 2.1972
    y = 1.0986
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    Re: Is this correct?

    ?
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  5. #5
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    Re: Is this correct?

    Quote Originally Posted by Mollers View Post
    thank you I got it!

    I feel dumb asking some of these because they look so basic but I want to make sure I know what I'm doing.

    1. The domain and range of 5x cubed + 4
    I got negative infinite - positive infinite.
    You mean both domain and range are negative to positive infinity (or "all real numbers"). Yes, that is correct.

    2. Write as exponential form: ln 1 = 0
    I got 10^0 = 1
    Yes.

    3. Use properties of logarithms to rewrite the expression: 4 log base 2 a squared
    I got log base 2 a^8
    Yes, that is correct. "8 log base 2 a" would also be a correct answer.

    4. Solve the equation: log x = -2
    I got 10^-2 = .01
    X = .01
    Yes, assuming that the logarithm is base 10. In advanced mathematics log base 10 is used so seldom that "log" is often used to mean natural logarithm but since you use "ln" in the next problem, you are probably correct that "log" is base 10.

    5. 2 ln y = 9
    I got 2y = ln 9
    2y = 2.1972
    No, no , no , no!!! You can't just "take" a ln from one side and put it on the other!
    First step- either use the properties of logarithms to write [itex]2 ln y= ln y^2= ln 9[/itex] or divide both sides by 2 to get ln(y)= (1/2)ln(9)= ln(9^{1/2})= ln(3).
    Then use the fact that ln is "one to one": if ln a= ln b then a= b.

    y = 1.0986
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    Re: Is this correct?

    wow thank you thank you!
    this is the last one
    Is this correct?-math-problem.png
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  7. #7
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    Re: Is this correct?

    Quote Originally Posted by Mollers View Post
    thank you I got it!

    I feel dumb asking some of these because they look so basic but I want to make sure I know what I'm doing.

    1. The domain and range of 5x cubed + 4
    I got negative infinite - positive infinite.

    2. Write as exponential form: ln 1 = 0
    I got 10^0 = 1

    3. Use properties of logarithms to rewrite the expression: 4 log base 2 a squared
    I got log base 2 a^8

    4. Solve the equation: log x = -2
    I got 10^-2 = .01
    X = .01

    5. 2 ln y = 9
    I got 2y = ln 9
    2y = 2.1972
    y = 1.0986
    2. is wrong. The natural logarithm is a logarithm of base e, not base 10.
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