untchick04tm

2022-01-22

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

recoronarrv

Beginner2022-01-22Added 20 answers

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. Youll

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. Youll

reinosodairyshm

Beginner2022-01-23Added 36 answers

RizerMix

Expert2022-01-27Added 656 answers

One way to get an approximation of a solution is to use the Lambert W function if we note that $n(n-1)\approx (n-\frac{1}{2}{)}^{2}$ .
Start with
$n(n-1){3}^{n}=91854$
Taking the square root and dividing by $\sqrt[4]{3}$ , we get
$(n-\frac{1}{2}){\sqrt{3}}^{n-\frac{1}{2}}\approx \frac{\sqrt{9}1854}{\sqrt[4]{3}}$
Rewriting ${\sqrt{3}}^{n-\frac{1}{2}}$ and multiplying by $\frac{1}{2}\mathrm{log}(3)$ , we get
$\frac{1}{2}\mathrm{log}(3)(n-\frac{1}{2}){e}^{\frac{1}{2}\mathrm{log}(3)(n-\frac{1}{2})}\approx \frac{1}{2}\mathrm{log}(3)\frac{\sqrt{91854}}{\sqrt[4]{3}}$
Therefore,
$\frac{1}{2}\mathrm{log}(3)(n-\frac{1}{2})\approx (\frac{1}{2}\mathrm{log}(3)\frac{91854}{\sqrt[4]{3}})$
Evaluating, we get
$n\approx 6.99578$
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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