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    Not sure how to rearrange this equation to find the number of years. So far I've got:
    5446.25=5000x(1+(2.5/200))^2n

    I think I have to use this formula but not exactly sure, I don't know what In means either:
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    Help, merci :bump:
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    (Original post by Kangie)

    Not sure how to rearrange this equation to find the number of years. So far I've got:
    5446.25=5000x(1+(2.5/200))^2n

    I think I have to use this formula but not exactly sure, I don't know what In means either:
    Let n = number of "half years". Form an equation in n, and rearrange to solve. You will need to use logarithms! Ln is the natural logarithm, log to the base e.
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    (Original post by lizard54142)
    Let n = number of "half years". Form an equation in n, and rearrange to solve. You will need to use logarithms!
    I haven't been taught logarithms! We don't have that in the syllabus for maths studies I don't think, if it is then why the hell hasn't my teacher taught me it?!!? D:
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    (Original post by Kangie)
    I haven't been taught logarithms! We don't have that in the syllabus for maths studies I don't think, if it is then why the hell hasn't my teacher taught me it?!!? D:
    Well you should have been, because it is crucial for this question! You also could do it by "guessing" values of n and seeing which value is closest, you wouldn't need logarithms for that.
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    (Original post by lizard54142)
    Well you should have been, because it is crucial for this question! You also could do it by "guessing" values of n and seeing which value is closest, you wouldn't need logarithms for that.
    I just found the syllabus and it said: 'Not required:use of logarithms to find n, given the sum ofthe first n terms; sums to infinity.'
    I was wondering if this was a trial and error question, wasn't sure though. I guess it is! Thanks
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    (Original post by Kangie)
    I just found the syllabus and it said: 'Not required:use of logarithms to find n, given the sum ofthe first n terms; sums to infinity.'
    I was wondering if this was a trial and error question, wasn't sure though. I guess it is! Thanks
    Okay! So just guess values of n, if the calculation is greater than what you need take smaller n etc... ask your teacher about logarithms, they would give you an exact answer
 
 
 
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