Which of the following is the highest value?

Question

Question:

Find the highest value among $12^9$, $10^{11}$ and $11^{10}$.

I have seen problems like this, but they had surds, these are integers. Also, the LCM of $10$, $11$, $9$ $(990)$ is fairly large to alow a normal approach. There must be some other approach. I know this must be fairly simple, but I am not getting it.

Please show me the right way.

Thanks.

Answer 1

Try using the binomial theorem, perhaps:

$12^9=(11+1)^9= \dots$,

$11^{10}=(10+1)^{10}=\dots$

Answer 2

Let $f(x)=x\ln(21-x)$ for $8\le x\le 13$, $$f'(x)=\ln(21-x)-\frac{x}{21-x}=\frac{(21-x)\ln(21-x)-x}{21-x},\quad\quad\quad 8< x< 13$$ Observe $8\le 21-x \le 13 \Longrightarrow 2<\ln 8\le \ln(21-x) \le \ln 13<4$, then $$16<(21-x)\ln(21-x)<52,\quad\quad\quad 8\le x\le 13$$ In particular $f'(x)>0$ for $8< x< 13$, what means $f$ is increasing in $]8,13[$, thus \begin{align} f(9)& < &f(10) & < & f(11)\\ 9\ln(12)& < &10\ln(11) & < & 11\ln(10)\\ \exp[9\ln(12]& < &\exp[10\ln(11)] & < & \exp[11\ln(10)] \end{align} Therefore $12^9<11^{10}<10^{11}$.