Determing which term of the geometric sequence a number is equal to

Question

The question is: Which term of the geometric sequece $4,12,36,\ldots$ is equal to $78732.$

My process of working it out is:

$t_n= t_1\cdot r^{n-1}$

$78372 = 4\cdot3^{n-1}$

$78372 = 12^{n-1}$

From here onwards I am not sure on what to do. Like how do I go further into solving which term it is. Any steps or comments if I am on the right track or any help is appreciated

Answer 1

$$4\cdot3^{n-1}=78732$$ or $$3^{n-1}=19683$$ (here was your mistake) or $$n=1+\frac{\ln19683}{\ln3},$$ which gives $n=10.$

Also, it's better to learn that $$3^9=19683.$$

Answer 2

The general term of the progression is $a_n=4\cdot3^{n-1}$ with $n\geq1$. So, I have: $$4\cdot3^{n-1}=78732$$Posing $\alpha=n-1$, I obtain $\alpha=9$ and so $n=10$.

Answer 3

Note that $78732 = 4\cdot3^{n-1}$ does not imply $78732 = 12^{n-1},$ as you wrote. Instead, from the first equation, divide both sides by $4$ to get $19683= 3^{n-1}.$ This says that $19683$ is some power of $3.$ Multiply both sides by $3$ to give $$3×19683 = 3^n.$$ Now taking logs to base $3$ gives $$1+\log{19683}=n,$$ which you may work out on a calculator.

Answer 4

Without logarithms:

This is the same as finding $n$ for $\dfrac{78732}4=19683$ in $\dfrac44,\dfrac{12}4,\dfrac{36}4,\cdots=3^0,3^1,3^2,\cdots$

To find the right power, for efficiency we square recursively,

$$3^2=9,3^4=81,3^8=6561$$ and notice that $19683=3\cdot6561=3^9$. Done.