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Geometric sequences; solving algebraically for exponents

  • Thread starter trulyfalse
  • Start date
  • #1
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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?

Homework Statement


Given the following geometric sequences, determine the number of terms, n.
t1=5
r (common ratio)=3
tn=135

Homework 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

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!
 

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36
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?

Homework Statement


Given the following geometric sequences, determine the number of terms, n.
t1=5
r (common ratio)=3
tn=135

Homework 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

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!
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.
 
  • #3
35
0
I'll just attribute my obliviousness to sleep deprivation and inattentiveness... Anyways, thanks for help! :)
 

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