How to solve the following logarithmic equation: $n(n-1)3^{n} = 91854$

Question

Could somebody please help me solve this equation $n(n-1)3^{n} = 91854$

I can simplify it down to $n\cdot\log(3)+\log(n) + \log(n-1) = \log(91854)$, but I have no idea where to go from there...

Thanks for any help you can give me

p.s. I know the answer is n=7, but no idea how to get there

Answer 1

To Solve Questions of this Type

There is no algebraic method to solve equations like this. If the number on the right hand side had been 91853 instead then the answer can not be written in terms of common expressions (trigonometry, logs/exponentials, arithmetic, etc). Generally questions of this type have to be answered numerically - which could include a number of ways: trial and error, computer software, graphing, etc.

To Solve this Specific Question (or ones where the answer is an integer)

Try to factorize 91854. You'll find that:

$$91854=2\times3^8\times7$$

The $7$ stands out as compared to the other numbers. As we are assuming integer solutions it must be used one of the parts on the left or a combination of them.

Some are easily excluded as they clearly do not have integer solutions. E.g.: $3^n=7\times something$, $n\times3^n=7\times something$, $(n-1)3^n=7\times something$, etc.

The one which looks potentially like a solution is: $n(n-1)=7\times something$

So you can conclude that either $n=7$ or $n-1=7$. Checking easily reveals that the answer is $n=7$.

Answer 2

Just keep dividing by $3$ until you get a number that isn't divisible by $3$.

Answer 3

n<1 => F < 91854

n>=1 F is increasing => F - 91854 has no more then 1 root

Answer 4

One way to get an approximation of a solution is to use the Lambert W function if we note that $n(n-1)\approx\left(n-\frac12\right)^2$.

Start with $$ n(n-1)3^n=91854 $$ Taking the square root and dividing by $\sqrt[4]{3}$, we get $$ \left(n-\frac12\right)\sqrt3^{\,n-\frac12}\approx\frac{\sqrt{91854}}{\sqrt[4]3} $$ Rewriting $\sqrt3^{\,n-\frac12}$ and multiplying by $\frac12\log(3)$, we get $$ \frac12\log(3)\left(n-\frac12\right)e^{\frac12\log(3)\left(n-\frac12\right)}\approx\frac12\log(3)\frac{\sqrt{91854}}{\sqrt[4]3} $$ Therefore, $$ \frac12\log(3)\left(n-\frac12\right)\approx W\left(\frac12\log(3)\frac{\sqrt{91854}}{\sqrt[4]3}\right) $$ Evaluating, we get $$ n\approx6.99578 $$ If $n$ is supposed to be an integer, a good guess would be $n=7$. Trying $n=7$ shows that that is indeed the answer.

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An algorithm to compute $W(x)$ is given in this answer.