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    in question 6c. my steps are fine until the last step:
    N log7/8 (7/8)< log7/8 1/320
    N < 43.2 (3sf)

    However the correct answer is N > 43.2.
    Can someone explain why?
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    (Original post by alesha98)
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    in question 6c. my steps are fine until the last step:
    N log7/8 (7/8)< log7/8 1/320
    N < 43.2 (3sf)

    However the correct answer is N > 43.2.
    Can someone explain why?
    You're dividing by a negatie number.
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    (Original post by alesha98)
    Name:  c2 log question.png
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    in question 6c. my steps are fine until the last step:
    N log7/8 (7/8)< log7/8 1/320
    N < 43.2 (3sf)

    However the correct answer is N > 43.2.
    Can someone explain why?
    Your penultimate line is also incorrect.

    Can you post your working in full?
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    (Original post by Zacken)
    You're dividing by a negatie number.
    my full working for part c is:
    160 - (20(1-(7/8)^N)) / 1-7/8 < 0.5
    20 - 20(1-(7/8)^N) < 1/16
    320 - 320(1-(7/8)^N) < 1
    320(1-(7/8)^N) > 319
    1-(7/8)^N > 319/320
    -(7/8)^N > -1/320
    (7/8)^N < 1/320
    N log7/8 (7/8) < log7/8 1/320
    N < 43.2 (3sf)

    May you tell me where exactly I got wrong please?
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    (Original post by notnek)
    Your penultimate line is also incorrect.

    Can you post your working in full?
    yes i just did
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    (Original post by alesha98)
    yes i just did
    Do you understand what it means to multiply and divide an inequality by a negative number?

    I'll let Notnek take over, I'm not home right now and can't check your working thoroughly.
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    remember that some logarithms are negative ?

    so if you divide by a logarithm you may need to flip the arrow....
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    (Original post by the bear)
    remember that some logarithms are negative ?

    so if you divide by a logarithm you may need to flip the arrow....
    I dont know anything about that, can you explain it please?
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    (Original post by alesha98)
    I dont know anything about that, can you explain it please?
    for instance log0.8 is a negative number ( using base 10 )

    so if you have

    X(log0.8) > 15

    X < 15/log0.8

    if instead you have

    X(log8) > 15

    X > 15/log8
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    (Original post by the bear)
    for instance log0.8 is a negative number ( using base 10 )

    so if you have

    X(log0.8) > 15

    X < 15/log0.8

    if instead you have

    X(log8) > 15

    X > 15/log8
    thanks. So if i log a number that is smaller than 1, then it is a negative number ?
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    (Original post by alesha98)
    thanks. So if i log a number that is smaller than 1, then it is a negative number ?
    yes that is true for any base.
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    (Original post by the bear)
    for instance log0.8 is a negative number ( using base 10 )

    so if you have

    X(log0.8) > 15

    X < 15/log0.8

    if instead you have

    X(log8) > 15

    X > 15/log8
    but from what i did, N log7/8 (7/8) < log7/8 1/320, log7/8 (7/8) is 1, so why do i need to flip the inequality?
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    (Original post by notnek)
    Your penultimate line is also incorrect.

    Can you post your working in full?
    what happen if I log with a base that is smaller than 1?
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    (Original post by alesha98)
    but from what i did, N log7/8 (7/8) < log7/8 1/320, log7/8 (7/8) is 1, so why do i need to flip the inequality?
    Taking logs of base a where 0&lt;a&lt;1 reverses the inequality.


    If you take logs base 10 instead then things are clearer:

    \displaystyle \left(\frac{7}{8}\right)^N &lt; \frac{1}{320}

    \displaystyle N \log_{10} \left(\frac{7}{8}\right)&lt; \log_{10} \left(\frac{1}{320}\right)

    Now \log_{10} \left(\frac{7}{8}\right) &lt; 0 so we need to reverse the inequality when dividing:

    \displaystyle N &gt; \frac{ \log_{10} \left(\frac{1}{320}\right)}{\log  _{10} \left(\frac{7}{8}\right)}
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    [QUOTE=notnek;64219381]Taking logs of base a where 0&lt;a&lt;1 reverses the inequality.


    thankyou so much
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    (Original post by alesha98)
    Taking logs of base a where 0&lt;a&lt;1 reverses the inequality.
    To explain this a bit further :

    \displaystyle \log_{\frac{7}{8}} x is a strictly decreasing function i.e. it looks like this :



    So e.g. 2 &lt; 3 but \displaystyle \log_{\frac{7}{8}} 2 &gt; \log_{\frac{7}{8}} 3 as you can see in the graph.


    \log_{10} x is strictly increasing so you don't need to reverse the inequality sign when taking logs base 10.
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    (Original post by alesha98)

    thankyou so much
    I'd suggest always using log base 10 or the natural log for these questions. Don't do things like logs with fractional bases or whatever, it's just complicating your life.
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    (Original post by Zacken)
    I'd suggest always using log base 10 or the natural log for these questions. Don't do things like logs with fractional bases or whatever, it's just complicating your life.
    I agree with this.

    Or you could treat a log inequality as an equation and only deal with the inequality sign at the end, similar to how one may tackle a quadratic inequality.
 
 
 
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