Solving $1.1^n = n^{100}$

Question

How do I go about solving for $n$ in the following equation: $$(1.1)^n = n^{100}$$ A hint suffices.

Answer 1

Taking logarithms, you see that this equation takes the form $an = b\log n$, or equivalently, $\dfrac{\log n}{n} = k$. For general values of $k$ (here $k=\dfrac{\log 1.1}{100}$), this solution has no neat expression since the function $n \mapsto \dfrac{\log n}{n}$ has no 'nice' inverse (multi-valued or otherwise), i.e. one expressible in terms of elementary functions.

However, it can be solved numerically using one of a large array of techniques.

Added: If it's any consolation, you can determine the number of solutions using analytic techniques, so that you're not led on a wild goose chase. To do this, you can show that the function $x \mapsto \dfrac{\log x}{x}$ takes each value $k$ with $0<k<\frac{1}{e}$ exactly twice on the real line: once in the interval $(1,e)$, and once in the interval $(e, \infty)$.

Answer 2

Take the log of each side, then plot the graph or solve numerically.

Answer 3

Algebraically, you can't isolate this equation for $n$. Solving numerically in mathematica, the solutions set to five decimal places is $\{1.00095, -0.999048, 9623.25\}$.

Answer 4

How about taking the log? You get $n\ln(1.1)=100\ln(n)$ or $\frac{n}{\ln(n)}=\frac{100}{ln(1.1)}$ which you can solve numerically.

Answer 5

Maple shows three real solutions and many,many complex solutions in terms of the Lambert W function. Real solutions: $$ \begin{align} \frac{-100 W \left(\frac{\operatorname{ln}(11/10)}{100}\right)}{\operatorname{ln} (11/10)} &= -0.9990482585 \\ \frac{-100W \left(\frac{-\operatorname{ln} (11/10)}{100}\right)}{\operatorname{ln} (11/10)} &= 1.000954467 \\ \frac{-100 W_{-1} \left(\frac{-\operatorname{ln} (11/10)}{100}\right)}{\operatorname{ln} (11/10)} &= 9623.250526 \end{align}$$

Answer 6

First you can rewrite the equation a bit, as

$$\begin{align}1.1^n &= n^{100} \\ 1.1^{n/100} &= n \\ (1.1^{1/100})^n &= n \\ p^n &= n & (\text{where } p = \sqrt[100]{1.1}).\end{align}$$

This is where you normally get stuck, but by introducing the somewhat common notation of the Lambert W function $W$ satisfying $z = W(z) e^{W(z)}$ you can "cheat" your way to a closed-form expression for $n$. Example 1 on the Wikipedia page shows how to get to the following solution:

$$n = \frac{W(-\log a)}{-\log a}$$

If you insist, you can find numerics in Mathematica with the command ProductLog, which is just a different name for the Lambert W function. Feeding it the above, you get the solutions $1.00095$ and $9623.25$.

Answer 7

Okay so we start with equation:

$$1.1^{n}=n^{100}$$

we take logarithms of both sides:

$$n\ln 1.1=100\ln n$$

Now taking squared difference of sides gives us a parabola:

$$y=(n\ln 1.1-100\ln n)^{2}$$

The parabola have minimum when derivative its is zero. However, the derivative still contains $\log n$ and this means the expression is truly non-linear.

From my knowledge, I can only take Taylor expansion of $y(n)$ to solve for step from initial guess. The numerical solution can then be written as:

$$lim_{i\to \infty}n_{i}$$

where

$$n_{i}=n_{i - 1}-(y'')^{-1}y'$$

and $n_{0}=1$ (or any other reasonable guess). Note that $(y'')^{-1}$ denotes inverse function of second derivative of $y$.