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Geometric sequences; solving algebraically for exponents

  1. Nov 24, 2011 #1
    Hello everyone! My question is twofold. Firstly, how do I solve for term numbers in a geometric sequence and secondly, how do I algebraically solve for variables that are exponents?

    1. The problem statement, all variables and given/known data
    Given the following geometric sequences, determine the number of terms, n.
    t1=5
    r (common ratio)=3
    tn=135

    2. Relevant equations
    tn=t1rn-1
    where t1 is the first term
    n is the number of terms
    r is the common ratio
    tn is the general term

    3. The attempt at a solution
    I substituted in all known values yielding 135=(5)(3)n-1, which I simplified to 27=(3)n-1, leaving me stuck at the exponent variable. From here, how do I algebraically solve for the variable? Thanks!
     
  2. jcsd
  3. Nov 24, 2011 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Well, 27 is a power of 3, so the answer is obvious by inspection. However, if you wanted to solve for it, a general way would be to use logarithms:

    log10(27) = log10(3n-1)

    and by the properties of logarithms, this becomes:

    log10(27) = (n-1)log10(3)

    n-1 = log10(27)/log10(3)

    Actually, since both sides of the equation were powers of 3, it would have made much more sense to take the base 3 logarithm of both sides, rather than the base 10 logarithm. However, your calculator probably doesn't compute base 3 logarithms.
     
  4. Nov 24, 2011 #3
    I'll just attribute my obliviousness to sleep deprivation and inattentiveness... Anyways, thanks for help! :)
     
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