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    Show that n =  \frac{log2000003}{log3}-1 from  \frac{3(1-3^n)}{1-3} = 1000000 without expanding the brackets. I'm not sure how to do this, thanks.
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    You have to expand the brackets to get n on its own
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    (Original post by Daiblain)
    Show that n =  \frac{log2000003}{log3}-1 from  \frac{3(1-3^n)}{1-3} = 1000000 without expanding the brackets. I'm not sure how to do this, thanks.
    Is your denominator supposed to be 1-3 or something else?
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    (Original post by aoxa)
    Is your denominator supposed to be 1-3 or something else?
    It's supposed to be 1-3

    (Original post by Suprite)
    You have to expand the brackets to get n on its own
    Can't you multiply the equation by -2 and divide by 3 to get n on its own? Isn't this able to be done this way as well?
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    (Original post by Daiblain)
    It's supposed to be 1-3



    Can't you multiply the equation by -2 and divide by 3 to get n on its own? Isn't this able to be done this way as well?
    Weird way to write the denominator, but yes, multiplying by -2 and dividing by 3 and then using log laws should get you the right answer.
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    (Original post by aoxa)
    Weird way to write the denominator, but yes, multiplying by -2 and dividing by 3 and then using log laws should get you the right answer.
    Yeah, sorry. It's a sum of a geometric sequence question.
    I already tried doing what you said and I couldn't get anywhere.
    I got to  1-3^n = -2000000/3
    bringing over the one and dividing by -1 gives  3^n = 2000000/3 +1
    taking logs then dividing by the log on the left side gives  n = [log\frac{2000000}{3} +log1]/log3
    Where have I went wrong? If im not wrong, is this just another way to express n where I'm not able to get to the expression that I'm looking for?

    (edited)
    Just realised 1 is the same as 3/3 so that 2000003/3 can be done, but I've lost the -1 somewhere. R.I.P - my algebra skills
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    (Original post by Daiblain)
    Yeah, sorry. It's a sum of a geometric sequence question.
    I already tried doing what you said and I couldn't get anywhere.
    I got to  1-3^n = -2000000/3
    bringing over the one and dividing by -1 gives  3^n = 2000000/3 +1
    taking logs then dividing by the log on the left side gives  n = [log\frac{2000000}{3} +log1]/log3
    Where have I went wrong? If im not wrong, is this just another way to express n where I'm not able to get to the expression that I'm looking for?

    (edited)
    Just realised 1 is the same as 3/3 so that 2000003/3 can be done, but I've lost the -1 somewhere. R.I.P - my algebra skills

    So 200003/3 = 3^n, taking log base 3 of each side gives log(3) 2000003/3 = n, and then using the log law about divisibility - log a/b = loga - logb so log(3)20000003 -log(3)3 = n, and then there's the log law of log(a)a = 1 which is where your -1 comes from, so you end up with log(3)200003 -1 =n.

    Hope that's not too bad to read, I can't be doing with latex at this time.
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    (Original post by aoxa)
    So 200003/3 = 3^n, taking log base 3 of each side gives log(3) 2000003/3 = n, and then using the log law about divisibility - log a/b = loga - logb so log(3)20000003 -log(3)3 = n, and then there's the log law of log(a)a = 1 which is where your -1 comes from, so you end up with log(3)200003 -1 =n.

    Hope that's not too bad to read, I can't be doing with latex at this time.
    But the question wants it in terms of base 10 and the log of 2000003 should be divided by the log of 3. It's fine, i'll just remember to expand the brackets every time i see a log question. This will bug me forever, though. Thanks alot.
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    (Original post by Daiblain)
    But the question wants it in terms of base 10 and the log of 2000003 should be divided by the log of 3. It's fine, i'll just remember to expand the brackets every time i see a log question. This will bug me forever, though. Thanks alot.
    Ah sorry, thats me misreading the question. Take base ten then, so you've got 2000003/3 = 3^n, and then log(10)200003/3 = log(10)3^n, then using the division log law thing again you get log 2000003 - log 3 = log3^n, taking the log 3 to the right hand side and using the multiplying log law you get log 200003 = (n+1)log3 (and using the log law where you bring the powers down) and then divide by log 3, take across the 1 and you get the answer, just with a bit more fiddling about then using base 3!
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    (Original post by aoxa)
    Ah sorry, thats me misreading the question. Take base ten then, so you've got 2000003/3 = 3^n, and then log(10)200003/3 = log(10)3^n, then using the division log law thing again you get log 2000003 - log 3 = log3^n, taking the log 3 to the right hand side and using the multiplying log law you get log 200003 = (n+1)log3 (and using the log law where you bring the powers down) and then divide by log 3, take across the 1 and you get the answer, just with a bit more fiddling about then using base 3!
    That clears it up. Thanks
 
 
 
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