Geometric sequences; solving algebraically for exponents

[trulyfalse]

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
t_n=135

Homework Equations

t_n=t₁r^n-1
where t₁ is the first term
n is the number of terms
r is the common ratio
t_n 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!

 

[cepheid]

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
t_n=135

Homework Equations

t_n=t₁r^n-1
where t₁ is the first term
n is the number of terms
r is the common ratio
t_n 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(3^n-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.

 

[trulyfalse]

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