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Math Help - 3 math problems.

  1. #1
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    3 math problems.

    Hi...
    I tried to solve them, but ended with nothing :-(



    Please Help!

    Thank you!!!
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by rectangle View Post
    Hi...
    I tried to solve them, but ended with nothing :-(



    Please Help!

    Thank you!!!


    you realize that

      5^4 = 625

    so you have

     625 ^x = 5^{4x} and  25^x = 5^{2x} and  125^x = 5^{3x}

    does this help you in any way ? show what have you try to do
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  3. #3
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    Quote Originally Posted by rectangle View Post
    Hi...
    I tried to solve them, but ended with nothing :-(

    ...

    Please Help!

    Thank you!!!
    To #1:

    1. Probably you have noticed that 5^4 =625 , 5^3=125, 5^2 = 25

    2. Your equation becomes:

    5^{4x}+3+6\cdot 5^{2x} = -4\cdot 5^x+2-4\cot 5^{3x}

    3. Rearrange the equation:

    5^{4x}+4\cot 5^{3x}+6\cdot 5^{2x} +4\cdot 5^x+1= 0

    4. This is a complete polynom of 4th degree:

    (5^x+1)^4=0

    This equation doesn't have any real solution.
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  4. #4
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    Hello, rectangle!

    \text{Solve: }\;81^x - 12\cdot27^x + 47\cdot9^x \;=\;72\cdot3^x - 36

    We have: . (3^4)^x - 12(3^3)^x + 47(3^2)^x - 72(3)^x + 36 \;=\;0

    . . . . . . . . (3^x)^4 - 12(3^x)^3 + 47(3^x)^2 - 72(3^x) + 36 \;=\;0


    Let u \,=\,3^x\!:\;\;u^4 - 12u^3 + 47u^2 - 72u + 36 \;=\;0


    Factor: . (u-1)(u-2)(u-3)(u-6) \;=\;0

    . . Hence: . u\;=\;1,\:2,\:3,\:6


    Back-substitute:

    . . \begin{array}{ccccccc}<br />
3^x \;=\; 1 & \Rightarrow  & x \;=\; 0 \\<br />
3^x \;=\; 2 & \Rightarrow & x \;=\; \log_32 \\<br />
3^x \;=\; 3 & \Rightarrow & x \;=\; 1 \\<br />
3^x \;=\; 6 & \Rightarrow & x \;=\; \log_36 \end{array}




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  5. #5
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    Quote Originally Posted by rectangle View Post
    Hi...
    I tried to solve them, but ended with nothing :-(

    ...
    Please Help!

    Thank you!!!
    To #3.1:

    2\left(\log_9(x)\right)^2=1+\log_9(x)

    Use the substitution u = \log_9(x). Your equation becomes:

    2u^2=1+u. Solve for u.

    Afterwards re-substitute and solve for x. I've got x = 9 ~\vee~x=\frac13
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  6. #6
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    Quote Originally Posted by yeKciM View Post
    you realize that

      5^4 = 625

    so you have

     625 ^x = 5^{4x} and  25^x = 5^{2x} and  125^x = 5^{3x}

    does this help you in any way ? show what have you try to do
    Thank you. It did help me a lot!

    Quote Originally Posted by earboth View Post
    To #1:

    1. Probably you have noticed that 5^4 =625 , 5^3=125, 5^2 = 25

    2. Your equation becomes:

    5^{4x}+3+6\cdot 5^{2x} = -4\cdot 5^x+2-4\cot 5^{3x}

    3. Rearrange the equation:

    5^{4x}+4\cot 5^{3x}+6\cdot 5^{2x} +4\cdot 5^x+1= 0

    4. This is a complete polynom of 4th degree:

    (5^x+1)^4=0

    This equation doesn't have any real solution.
    Thank you so much.

    Quote Originally Posted by earboth View Post
    To #3.1:

    2\left(\log_9(x)\right)^2=1+\log_9(x)

    Use the substitution u = \log_9(x). Your equation becomes:

    2u^2=1+u. Solve for u.

    Afterwards re-substitute and solve for x. I've got x = 9 ~\vee~x=\frac13
    Thank you. Can I solve the second one (3.2) also in that way?

    Quote Originally Posted by Soroban View Post
    Hello, rectangle!


    We have: . (3^4)^x - 12(3^3)^x + 47(3^2)^x - 72(3)^x + 36 \;=\;0

    . . . . . . . . (3^x)^4 - 12(3^x)^3 + 47(3^x)^2 - 72(3^x) + 36 \;=\;0


    Let u \,=\,3^x\!:\;\;u^4 - 12u^3 + 47u^2 - 72u + 36 \;=\;0


    Factor: . (u-1)(u-2)(u-3)(u-6) \;=\;0

    . . Hence: . u\;=\;1,\:2,\:3,\:6


    Back-substitute:

    . . \begin{array}{ccccccc}<br />
3^x \;=\; 1 & \Rightarrow  & x \;=\; 0 \\<br />
3^x \;=\; 2 & \Rightarrow & x \;=\; \log_32 \\<br />
3^x \;=\; 3 & \Rightarrow & x \;=\; 1 \\<br />
3^x \;=\; 6 & \Rightarrow & x \;=\; \log_36 \end{array}


    Thank you. I did not understand the "Factor" Part. It would be great if someone could explain me in more details.


    Thank you all
    I feel like an idiot.
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