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Thread: [ASK]Logarithmic Inequation

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    [ASK]Logarithmic Inequation

    How to solve this?
    $\displaystyle (log_2x)^2+4>5log_7x+log_3x^2$
    This is what I've done so far (attached):

    How to solve the quadratic inequation formed? Factorization is an obvious no, and the quadratic formula will just make everything messier, is there any other way?
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    Re: [ASK]Logarithmic Inequation

    Quote Originally Posted by Monoxdifly View Post
    How to solve this?
    $\displaystyle (log_2x)^2+4>5log_7x+log_3x^2$
    This is what I've done so far (attached):
    How to solve the quadratic inequation formed? Factorization is an obvious no, and the quadratic formula will just make everything messier, is there any other way?
    I would change to the standard base: $\log_2(x)=\dfrac{\log(x)}{\log(2)},~\log_7(x)$$= \dfrac{\log(x)}{\log(7)},~\&~\log_3(x^2)= \dfrac{\log(x^2)}{\log(3)}$
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    Re: [ASK]Logarithmic Inequation

    Assume $x>1$

    Let
    $\displaystyle a=\log _2x$

    $\displaystyle b=\log _3x$

    $\displaystyle c=\log _7x$

    $\displaystyle 2^a=3^b=7^c$

    Show $b<a$ and $5c<2a$

    and therefore

    $\displaystyle 5c+2b<4a\leq a^2+4$
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    Re: [ASK]Logarithmic Inequation

    Quote Originally Posted by Plato View Post
    I would change to the standard base: $\log_2(x)=\dfrac{\log(x)}{\log(2)},~\log_7(x)$$= \dfrac{\log(x)}{\log(7)},~\&~\log_3(x^2)= \dfrac{\log(x^2)}{\log(3)}$
    I did that before submitting the one I posted in the attachment, and it didn't help much if at all.

    Quote Originally Posted by Idea View Post
    Show $b<a$ and $5c<2a$

    and therefore

    $\displaystyle 5c+2b<4a\leq a^2+4$
    Where did that 5c < 2a come from?
    And also, how did you get the last inequation?
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    Re: [ASK]Logarithmic Inequation

    Where did that 5c < 2a come from?

    it comes from $$2^5<7^2$$

    And also, how did you get the last inequation?[/QUOTE]

    $4a\leq a^2+4$ comes from $(a-2)^2 \geq 0$
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    Re: [ASK]Logarithmic Inequation

    I think I am having a hard time comprehending this...
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    Re: [ASK]Logarithmic Inequation

    More detailed explanation

    $$2^5<7^2$$
    $$2^{5c}<7^{2c}=2^{2a}$$
    $$5c<2a$$
    $$2b<2a$$

    Add

    $$5c+2b<4a\leq a^2+4$$
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    Re: [ASK]Logarithmic Inequation

    Okay, I understand the 5c + 2b < 4a part now, but why do we need that a + 4? You said that it came from (a – 2), but where did we get that from?
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