# Geometric sequences; solving algebraically for exponents

• trulyfalse
In summary, to solve for term numbers in a geometric sequence, substitute in known values and simplify. To algebraically solve for a variable in an exponent, use logarithms.

#### 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
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!

trulyfalse said:
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.

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

## What is a geometric sequence?

A geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a constant number called the common ratio. For example, in the sequence 2, 6, 18, 54, 162, the common ratio is 3.

## How do you find the common ratio in a geometric sequence?

To find the common ratio in a geometric sequence, you divide any term by the previous term. The resulting quotient will be the common ratio. For example, in the sequence 2, 6, 18, 54, 162, the common ratio is 6/2 = 3.

## How do you find the nth term in a geometric sequence?

The formula for finding the nth term in a geometric sequence is a_n = a_1 * r^(n-1), where a_n is the nth term, a_1 is the first term, and r is the common ratio. For example, in the sequence 2, 6, 18, 54, 162, the first term is 2, the common ratio is 3, and the 5th term is a_5 = 2 * 3^(5-1) = 2 * 3^4 = 162.

## How do you solve for the exponent in a geometric sequence?

To solve for the exponent in a geometric sequence, you can use logarithms. Take the logarithm of both sides of the equation a_n = a_1 * r^(n-1) and solve for n. For example, in the sequence 2, 6, 18, 54, 162, to find the exponent that would result in a term of 972, we can take the logarithm of both sides and solve for n: log(972) = log(2) + n * log(3). This gives us n = log(972/2)/log(3) = 4.

## What is the difference between a finite and infinite geometric sequence?

A finite geometric sequence has a limited number of terms, while an infinite geometric sequence has an unending number of terms. In a finite sequence, the terms will eventually stop increasing or decreasing, while in an infinite sequence, the terms will continue to increase or decrease indefinitely.