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Math Help - check work2

  1. #1
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    check work2

    1. if  3^{2x-3}=27

    x=2

    2. if 2^{x-1}= \frac{1}{16}

    x=5

    3. if 2^{x+2}-2^x=12<br />

    x=-2

    4. if 300(1.05)^n=600

    n=2.78

    5. to solve 3^{x+2}-3^x=216 you must use the logarithmic method
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  2. #2
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    Quote Originally Posted by william View Post
    1. if  3^{2x-3}=27

    x=2 <<< No

    2. if 2^{x-1}= \frac{1}{16}

    x=5 <<< No

    3. if 2^{x+2}-2^x=12<br />

    x=-2 <<< No


    4. if 300(1.05)^n=600

    n=2.78 <<< No


    5. to solve 3^{x+2}-3^x=216 you must use the logarithmic method
    How did you do these calculations?

    Obviously you need a lot of help. But we must know where and why you've made the mistakes. Otherwise we can't help you sufficiently.

    I'll take #5:

    3^{x+2}-3^x=216~\implies~9\cdot 3^{x}-3^x=216 ~\implies~8 \cdot 3^x=216 ~\implies~3^x=27~\implies~x=3
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  3. #3
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    Quote Originally Posted by earboth View Post
    How did you do these calculations?

    Obviously you need a lot of help. But we must know where and why you've made the mistakes. Otherwise we can't help you sufficiently.

    I'll take #5:

    3^{x+2}-3^x=216~\implies~9\cdot 3^{x}-3^x=216 ~\implies~8 \cdot 3^x=216 ~\implies~3^x=27~\implies~x=3
    new answers

    for 1. x=2
    2. x=-3
    3. x=2
    4. n=0.30
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  4. #4
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    Quote Originally Posted by william View Post
    1. if  3^{2x-3}=27

    x=2

    2. if 2^{x-1}= \frac{1}{16}

    x=5

    3. if 2^{x+2}-2^x=12<br />

    x=-2

    4. if 300(1.05)^n=600

    n=2.78

    5. to solve 3^{x+2}-3^x=216 you must use the logarithmic method
    Quote Originally Posted by william View Post
    new answers

    for 1. x=2
    2. x=-3
    3. x=2
    4. n=0.30
    No, x still isn't 2.
    In this case note that 27 = 3^3

    2. Yes

    3. Yup

    4. nay

    Divide both sides through by 300:

    1.05^n = 2
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  5. #5
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    Quote Originally Posted by william View Post
    new answers

    for 1. x=2
    ...
    You can check yourself if you've got the correct solution:

    Plug in the value of x into the original equation. If you get the same value on both sides of the equation your solution is OK. If not then ...

    3^{2x-3}=27 plug in x = 2 (that's your solution!):

    3^{2\cdot 2-3}=27~\implies~3^1 = 27~\implies~3 \neq 27

    Thus your result is wrong.
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  6. #6
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    Quote Originally Posted by e^(i*pi) View Post
    No, x still isn't 2.
    In this case note that 27 = 3^3

    2. Yes

    3. Yup

    4. nay

    Divide both sides through by 300:

    1.05^n = 2
    so 1. x=3

    4. x=14.21
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  7. #7
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    Quote Originally Posted by william View Post
    so 1. x=3

    4. x=14.21
    To check do what earboth suggested in post 5 but I get those from my calculations
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