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Math Help - Can you help me solve these so I can prepare for a test?

  1. #1
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    Can you help me solve these so I can prepare for a test?

    Most of these I can't do so I really need the answer to see how you did it in reverse. If you can't solve all of them I'd appreciate anything you can do to help.

    1. If is reflected in the y-axis state what would its equation be:

    ?

    2. Evaluate the following:



    no idea

    3. What is the inverse of:

    ?

    4. Solve:

    a)
    no idea

    b)





    ?

    c)

    once again have no idea

    5. Convert the following: into the form (if this makes no sense i might have typed it wrong)

    EDIT: One more for now:

    Solve: Can you help me solve these so I can prepare for a test?-codecogseqn.gif and express the answer correct to three decimal places.

    I can attempt this one but im not sure how to calculate it
    Can you help me solve these so I can prepare for a test?-codecogseqn-1-.gif
    Last edited by Rayzala; July 19th 2014 at 04:43 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Can you help me solve these so I can prepare for a test?

    Too many questions for a single thread. In the future no more than three.


    Quote Originally Posted by Rayzala View Post
    1. If is reflected in the y-axis state what would its equation be:

    ?
    y = 1/4 is not a typical function to graph. Is there an x missing in here somewhere?

    Quote Originally Posted by Rayzala View Post
    2. Evaluate the following:



    no idea
    500 = 5 x 10^2. So log(500) = ?

    Quote Originally Posted by Rayzala View Post
    3. What is the inverse of:

    ?
    To get an inverse you switch x and y. So if you have y = 2^{x - 1} then you needs to solve x = 2^{y - 1} for y. How do you get the y - 1 out of the exponent?

    Try that out for now.

    -Dan
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    Re: Can you help me solve these so I can prepare for a test?

    Hello, Rayzala!


    \text{4(a) Solve: }\:2^{x-2} + 2^x \:=\:320

    Factor: . 2^{x-2}(1 + 2^2) \:=\:320 \quad\Rightarrow\quad 5\cdot2^{x-2} \:=\:320 \quad\Rightarrow\quad 2^{x-2} \:=\:64

    . . . . . . 2^{x-2} \:=\:2^6 \quad\Rightarrow\quad x-2 \:=\:6 \quad\Rightarrow\quad x \:=\:8




    \text{4(b) Solve: }\:\log_2(3x+1) \:=\:4

    3x+1 \:=\:2^4 \quad\Rightarrow\quad 3x+1 \:=\:16 \quad\Rightarrow\quad 3x \:=\:15 \quad\Rightarrow\quad x \:=\:5




    \text{5. Convert the following: }\: y\:=\:\log_2(8x-16)^2

    y \;=\;\log_2\big[8(x-2)\big]^2 \;=\;\log_2\big[2^3(x-2)\big]^2  \;=\;\log_2\big[(2^3)^2(x-2)^2\big]

    y \;=\;\log_2\big[(2^6)(x-2)^2\big] \;=\;\log_2(2^6) + \log_2(x-2)^2 \;=\;6 + 2\log_2(x-2)




    \text{6. Solve to 3 decimal places: }\:3^x \:=\:300

    Take logs: . \ln(3^x) \:=\:\ln(300) \quad\Rightarrow\quad x\ln(3) \:=\:\ln(300)

    . . . . . . . . . x \;=\;\frac{\ln(300)}{\ln(3)} \;=\;5.191806549 \;\approx\;5.192
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    Re: Can you help me solve these so I can prepare for a test?

    Quote Originally Posted by topsquark View Post
    Too many questions for a single thread. In the future no more than three.

    Sorry Dan you were right, put a 2 instead of an x.. it should read: Can you help me solve these so I can prepare for a test?-codecogseqn-2-.gif Would it be: Can you help me solve these so I can prepare for a test?-codecogseqn-3-.gif? You completely lost me with the 2nd one though.. not sure how to continue from 'log(500) ='

    as for the third one... to get y-1 out of the exponent wouldn't I square both sides?`But I don't know how to get to the final answer. (Edit: turns out i read the question wrong and i was just supposed to graph the inverse which is much easier.. dont really need the equation but if you can help me i'd like to see how anyway)


    I will make sure to make another thread for the other questions because you said there's too many here. Thanks to the both of you.
    Last edited by Rayzala; July 20th 2014 at 07:52 AM.
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    Re: Can you help me solve these so I can prepare for a test?

    Quote Originally Posted by Rayzala View Post
    Sorry Dan you were right, put a 2 instead of an x.. it should read: Click image for larger version. 

Name:	CodeCogsEqn (2).gif 
Views:	12 
Size:	552 Bytes 
ID:	31315 Would it be: Click image for larger version. 

Name:	CodeCogsEqn (3).gif 
Views:	12 
Size:	568 Bytes 
ID:	31316? You completely lost me with the 2nd one though.. not sure how to continue from 'log(500) ='

    as for the third one... to get y-1 out of the exponent wouldn't I square both sides?`But I don't know how to get to the final answer.


    I will make sure to make another thread for the other questions because you said there's too many here. Thanks to the both of you.
    Absolutely not! The inverse of an exponential function is a logarithmic function. First, I would simplify by writing the function as $\displaystyle \begin{align*} y = 2^{-x} \end{align*}$. Now swap the x and y values and solve for y.
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    Re: Can you help me solve these so I can prepare for a test?

    Quote Originally Posted by Rayzala View Post
    Click image for larger version. 

Name:	CodeCogsEqn (2).gif 
Views:	12 
Size:	552 Bytes 
ID:	31315
    To reflect a set of points over the y-axis means that your are changing x to -x. So you are correct!

    Quote Originally Posted by Rayzala View Post
    [ATTACH=CONFIG]
    log_4(500) = log_4(5 \cdot 10^2) = log_4(5) + log_4(10^2) = log_4(5) + 2 \cdot log_4(10)

    Can you finish from here?

    -Dan
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