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Math Help - check work

  1. #1
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    check work

    1. the first step to solve the equation log_5(x+2)=2 is to take logs with base 2 of both sides

    2. the first step in solving the equation 2^x-2^{x+2}=8 is divide all terms by 2

    3. the first error in the solved equation
    \begin{array}{l}log(x+5)=2\\<br />
log(x+5)=log2\\<br />
x+5=2\\<br />
x=7\end{array}

    is taking the log of 2

    4. the next step in the partially solved equation
    \begin{array}{l}<br />
30(1.02)^n=600\\<br />
(1.02)^n=20\end{array}

    is bring n infront of the 1.02
    Last edited by Chris L T521; July 20th 2009 at 09:37 AM. Reason: improved on existing LaTeX
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by william View Post
    1. the first step to solve the equation log_5(x+2)=2 is to take logs with base 2 of both sides
    Hint:
    Spoiler:
    2=2\cdot 1 =2\log_5 5=\log_5 25
    So \log_5(x+2)=\log_5 25\implies\dots\implies x=\dots


    2. the first step in solving the equation 2^x-2^{x+2}=8 is divide all terms by 2
    No. Note that 2^{x+2}=2^2\cdot 2^x=4\cdot 2^x. Thus 2^x-2^{x+2}=2^x-2\cdot 2^x=-3\cdot 2^x=8. Do you think this has a solution?

    3. the first error in the solved equation
    \begin{array}{l}log(x+5)=2\\<br />
log(x+5)=log2\\<br />
x+5=2\\<br />
x=7\end{array}

    is taking the log of 2
    Very good. What would be the correct first step? (See my spoiler response to your first problem and follow the same idea...)

    4. the next step in the partially solved equation
    \begin{array}{l}<br />
30(1.02)^n=600\\<br />
(1.02)^n=20\end{array}

    is bring n infront of the 1.02
    Yes, but how do you bring the n out in front?
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  3. #3
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    so for 1. would separation of log_5(x+2) to log_5x+log_52 work?

    and 2. combine 2^x-2^{x+2} into 2^{x+2}
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  4. #4
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    Quote Originally Posted by william View Post
    so for 1. would separation of log_5(x+2) to log_5x+log_52 work?

    and 2. combine 2^x-2^{x+2} into 2^{x+2}
    1&2. No it would not

    You can only use those laws when the terms are multiplied or divided

    1 is fine as is

    2 could be put as:
    2^x - 4 \times 2^x = -3 \times 2^x
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    1&2. No it would not

    You can only use those laws when the terms are multiplied or divided

    1 is fine as is

    2 could be put as:
    2^x - 4 \times 2^x = -3 \times 2^x
    for 1 can you put it into exponential form?

    for 2 could you write 8 as 2^3 or separate (2)^x(2)^2
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  6. #6
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    Quote Originally Posted by william View Post
    for 1 can you put it into exponential form?

    for 2 could you write 8 as 2^3 or separate (2)^x(2)^2
    1. If you like. If f(x) = log_5{(x+2)} then x+2 = 5^{f(x)} and x = 5^{f(x)} - 2

    2. You can write 8 as 2^3. I broke down 2^{x+2} into 2^22^x, it is just more usual to simplify 2^2 to 4.
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    1. If you like. If f(x) = log_5{(x+2)} then x+2 = 5^{f(x)} and x = 5^{f(x)} - 2

    2. You can write 8 as 2^3. I broke down 2^{x+2} into 2^22^x, it is just more usual to simplify 2^2 to 4.
    for 2, which step would come first? writing 8 as 2^3 or breaking 2^{x+2} down into 2^22^x
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  8. #8
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    Quote Originally Posted by william View Post
    for 2, which step would come first? writing 8 as 2^3 or breaking 2^{x+2} down into 2^22^x
    It doesn't matter ^.^
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