Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
$1.1^1>1^{100}$ and $1.1^2<2^{100}$ hence one of the solutions is between $1$ and $2$. Another solution is much bigger – Hagen von Eitzen Sep 19 '12 at 17:51
@HagenvonEitzen How do you know the number of solutions? – saadtaame Sep 19 '12 at 17:57
I didn't say there were exactly 2 solutions. However, exponential grows faster than polynomial, hence for big an there will hold $(1.1)^n>n^{100}$ again. Haveing $n$ as unknown variablw when (apparently) no positiv einteger solution exists is a bit wierd, and it let me overlook that there must also be one negative solution. – Hagen von Eitzen Sep 19 '12 at 18:00
up vote 2 down vote accepted

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 $(e, \infty)$.

share|cite|improve this answer

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

share|cite|improve this answer
Can't it be solved algebraically? – saadtaame Sep 19 '12 at 18:00

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

share|cite|improve this answer
How do we prove it cannot be solved algebraically? – saadtaame Sep 19 '12 at 18:01
That's a good question; I actually don't know a proof. Perhaps investigate the Lambert W function link – sourisse Sep 19 '12 at 18:13

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.

share|cite|improve this answer

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

share|cite|improve this answer
I assumed that $1.1$ means $11/10$. – GEdgar Sep 19 '12 at 18:17

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 $9623.25$.

share|cite|improve this answer

Okay so we start with equation:


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:

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


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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.