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Math Help - More log stuff

  1. #1
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    More log stuff

    1. \log_{2}(25^(x+3) - 1) = 2 + \log_{2}(5 ^ (x+3) + 1)
    2. \log_{4}(2 \cdot 3^(x) - 6) -\log_{2}(9^(x) - 6)
    3. \log_{4}(2 \cdot 4^(x-2) - 1) + 4 = 2x

    I attempted the first to come as far as this-:
    Taking x+3 = n,
    25^(n) - 1/5^(n) + 1 =  4
    I've forgotten exponential mathematics it seems..
    Pardon me if I seem like a total noob, but I need help :P
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  2. #2
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    Hello,

    For latex use, do x^{...}, not x^(...)

    Quote Originally Posted by SVXX View Post
    1. \log_{2}(25^(x+3) - 1) = 2 + \log_{2}(5 ^ (x+3) + 1)
    2. \log_{4}(2 \cdot 3^(x) - 6) -\log_{2}(9^(x) - 6)
    3. \log_{4}(2 \cdot 4^(x-2) - 1) + 4 = 2x

    I attempted the first to come as far as this-:
    Taking x+3 = n,
    25^(n) - 1/5^(n) + 1 =  4
    I've forgotten exponential mathematics it seems..
    Pardon me if I seem like a total noob, but I need help :P
    Be careful with logs : \log (a+b) \neq \log a+ \log b

    \log_{2}(25^n - 1) = 2 + \log_{2}(5^n + 1)

    \log_2 (25^n -1)-\log_2 (5^n+1)=2

    Using the identity : \log a-\log b=\log \frac ab :

    \log_2 \frac{25^n-1}{5^n+1}=2

    \frac{25^n-1}{5^n+1}=4

    25^n-1=4 \cdot 5^n+4

    Substitute u=5^n
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  3. #3
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    I still didn't get it!
    25^{n} - 4\cdot u = 5
    5^{2n} - 4\cdot u = 5
    Have to get rid of that 2n somehow!
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  4. #4
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    Quote Originally Posted by SVXX View Post
    I still didn't get it!
    25^{n} - 4\cdot u = 5
    5^{2n} - 4\cdot u = 5
    Have to get rid of that 2n somehow!
    5^{2n}=(5^n)^2=u^2

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  5. #5
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    Alright, now I know for sure my head is sleeping.
    From that, x + 3 = 1, x = -2.
    Skipping the 2nd one, there's no equation to it...
    In the 3rd,
    \log_{4}(2 \cdot 4^{x-2} - 1) + \log_{4}256 = 2 \cdot x
    \log_{4}((2 \cdot 4^{x-2} -1) \cdot 256) = 2 \cdot x
    ((2 \cdot 4^{x-2} - 1) \cdot 256) = 4^{2x}
    Some help on this? Your advice comes really quick, thanks a lot for that
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  6. #6
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    Quote Originally Posted by SVXX View Post
    Alright, now I know for sure my head is sleeping.
    From that, x + 3 = 1, x = -2.


    Skipping the 2nd one, there's no equation to it...
    Yep, strange

    In the 3rd,
    \log_{4}(2 \cdot 4^{x-2} - 1) + \log_{4}256 = 2 \cdot x
    \log_{4}((2 \cdot 4^{x-2} -1) \cdot 256) = 2 \cdot x
    ((2 \cdot 4^{x-2} - 1) \cdot 256) = 4^{2x}
    Some help on this?
    Transform the powers..

    4^{x-2}=2^{2x-4}

    4^{2x}=2^{4x}=(2^{2x})^2

    --> (2^{2x-3}-1) \cdot 2^8=(2^{2x})^2

    2^{2x+5}-2^8=(2^{2x})^2

    2^5 \cdot 2^{2x}-2^8=(2^{2x})^2

    Is there a substitution you can make ?

    Your advice comes really quick, thanks a lot for that
    haha, you're welcome =)
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  7. #7
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    Quote Originally Posted by Moo View Post




    Yep, strange



    Transform the powers..

    4^{x-2}=2^{2x-4}

    4^{2x}=2^{4x}=(2^{2x})^2

    --> (2^{2x-3}-1) \cdot 2^8=(2^{2x})^2

    2^{2x+5}-2^8=(2^{2x})^2

    2^5 \cdot 2^{2x}-2^8=(2^{2x})^2

    Is there a substitution you can make ?



    haha, you're welcome =)
    Oh yeez, those substitutions
    2^{5} \cdot 2^{2x} - 2^{8} = (2 \cdot 2x)^2
    Taking 2 raised to 2x as u,
    u^{2} - 32 \cdot u + 256 = 0
    We get equal roots, u = 16.
     2^{2x} = 16
    2x = 4
    x = 2
    Thanks for the support!
    And don't mind if I come up with a newby sum again :P
    Vous Ítes les meilleurs! ^_^
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  8. #8
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    Quote Originally Posted by SVXX View Post
    Thanks for the support!
    And don't mind if I come up with a newby sum again :P
    Vous Ítes les meilleurs! ^_^
    It's not really newby oO
    All of this comes with training ^^

    Haha, good French speaking ^'
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  9. #9
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    I'm not so good at French, but still..:P
    Ok, the last problem. And this one isn't so sweet as the others were...

    |\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1
    There's a modulus sign...there are separate ways of solving modulus problems..I guess I have to integrate those ways.
    But yeah, better off is to ask you =P
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  10. #10
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    Quote Originally Posted by SVXX View Post
    I'm not so good at French, but still..:P
    Ok, the last problem. And this one isn't so sweet as the others were...

    |\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1
    There's a modulus sign...there are separate ways of solving modulus problems..I guess I have to integrate those ways.
    But yeah, better off is to ask you =P


    Differentiate the case when \frac{4-x}{2x-5}>1 (then |\log_{2}(\frac{4-x}{2x - 5})|=\log_{2}(\frac{4-x}{2x - 5})) and when \frac{4-x}{2x-5}<1 (then |\log_{2}(\frac{4-x}{2x - 5})|=-\log_{2}(\frac{4-x}{2x - 5}))

    I think it's quite tricky... Because you have to find the domain when it's <1 and when it's >1...

    It's the only way I see so far...
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  11. #11
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    Quote Originally Posted by Moo View Post


    Differentiate the case when \frac{4-x}{2x-5}>1 (then |\log_{2}(\frac{4-x}{2x - 5})|=\log_{2}(\frac{4-x}{2x - 5})) and when \frac{4-x}{2x-5}<1 (then |\log_{2}(\frac{4-x}{2x - 5})|=-\log_{2}(\frac{4-x}{2x - 5}))

    I think it's quite tricky... Because you have to find the domain when it's <1 and when it's >1...

    It's the only way I see so far...
    Yeah, there will be two cases here..
    If it is > 1, then
    x = \sqrt{10}
    If it is < 1, then
    x = 2.88, -1.38
    Discarding 2.88,
    x = -1.38
    Thats how it came via calculations...correct me if I'm wrong!
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  12. #12
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    Quote Originally Posted by SVXX View Post
    Yeah, there will be two cases here..
    If it is > 1, then
    x = \sqrt{10}
    (Or -\sqrt{10})

    It's \frac{4-x}{2x-5}>1, not x>1

    But after you've found the roots, you have to check if it's really >1.

    \frac{4-\sqrt{10}}{2x-5} \approx 0.632 <1

    So \sqrt{10} is not a solution.

    This goes the same way for -\sqrt{10}

    If it is < 1, then
    x = 2.88, -1.38
    Discarding 2.88,
    x = -1.38
    Once again, don't check if x<1, but if \frac{4-x}{2x-5}<1
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  13. #13
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    Quote Originally Posted by Moo View Post
    (Or -\sqrt{10})

    It's \frac{4-x}{2x-5}>1, not x>1

    But after you've found the roots, you have to check if it's really >1.

    \frac{4-\sqrt{10}}{2x-5} \approx 0.632 <1

    So \sqrt{10} is not a solution.

    This goes the same way for -\sqrt{10}



    Once again, don't check if x<1, but if \frac{4-x}{2x-5}<1
    In that case, x = -1.38!
    You're a femme fatale when it comes to maths
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  14. #14
    Moo
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    Quote Originally Posted by SVXX View Post
    In that case, x = -1.38!
    Yes

    But the most important thing : do you understand why ?


    You're a femme fatale when it comes to maths
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  15. #15
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    Quote Originally Posted by Moo View Post
    Yes

    But the most important thing : do you understand why ?




    Oh yez I know :P
    Next time I'll ask you directly when I get a problem lol
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