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Thread: More log stuff

  1. #1
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    More log stuff

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

    I attempted the first to come as far as this-:
    Taking x+3 = n,
    $\displaystyle 25^(n) - 1/5^(n) + 1 = 4$
    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.$\displaystyle \log_{2}(25^(x+3) - 1) = 2 + \log_{2}(5 ^ (x+3) + 1)$
    2.$\displaystyle \log_{4}(2 \cdot 3^(x) - 6) -\log_{2}(9^(x) - 6)$
    3.$\displaystyle \log_{4}(2 \cdot 4^(x-2) - 1) + 4 = 2x$

    I attempted the first to come as far as this-:
    Taking x+3 = n,
    $\displaystyle 25^(n) - 1/5^(n) + 1 = 4$
    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 : $\displaystyle \log (a+b) \neq \log a+ \log b$

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

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

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

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

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

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

    Substitute $\displaystyle u=5^n$
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  3. #3
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    I still didn't get it!
    $\displaystyle 25^{n} - 4\cdot u = 5$
    $\displaystyle 5^{2n} - 4\cdot u = 5$
    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!
    $\displaystyle 25^{n} - 4\cdot u = 5$
    $\displaystyle 5^{2n} - 4\cdot u = 5$
    Have to get rid of that 2n somehow!
    $\displaystyle 5^{2n}=(5^n)^2=u^2$

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

    $\displaystyle 4^{x-2}=2^{2x-4}$

    $\displaystyle 4^{2x}=2^{4x}=(2^{2x})^2$

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

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

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

    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..

    $\displaystyle 4^{x-2}=2^{2x-4}$

    $\displaystyle 4^{2x}=2^{4x}=(2^{2x})^2$

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

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

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

    Is there a substitution you can make ?



    haha, you're welcome =)
    Oh yeez, those substitutions
    $\displaystyle 2^{5} \cdot 2^{2x} - 2^{8} = (2 \cdot 2x)^2$
    Taking 2 raised to 2x as u,
    $\displaystyle u^{2} - 32 \cdot u + 256 = 0$
    We get equal roots, u = 16.
    $\displaystyle 2^{2x} = 16$
    $\displaystyle 2x = 4$
    $\displaystyle x = 2$
    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...

    $\displaystyle |\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1$
    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...

    $\displaystyle |\log_{2}(\frac{4-x}{2x - 5})| + \log_{2}x = 1$
    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 $\displaystyle \frac{4-x}{2x-5}>1$ (then $\displaystyle |\log_{2}(\frac{4-x}{2x - 5})|=\log_{2}(\frac{4-x}{2x - 5})$) and when $\displaystyle \frac{4-x}{2x-5}<1$ (then $\displaystyle |\log_{2}(\frac{4-x}{2x - 5})|=-\log_{2}(\frac{4-x}{2x - 5})$)

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

    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
    $\displaystyle x = \sqrt{10}$
    If it is < 1, then
    $\displaystyle x = 2.88, -1.38$
    Discarding 2.88,
    $\displaystyle x = -1.38$
    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
    $\displaystyle x = \sqrt{10}$
    (Or $\displaystyle -\sqrt{10}$)

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

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

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

    So $\displaystyle \sqrt{10}$ is not a solution.

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

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

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

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

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

    So $\displaystyle \sqrt{10}$ is not a solution.

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



    Once again, don't check if x<1, but if $\displaystyle \frac{4-x}{2x-5}<1$
    In that case, x = -1.38!
    You're a femme fatale when it comes to maths
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  14. #14
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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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