Which of the following numbers is greater? Without using a calculator and logarithm.
$$100^{101} , 101^{100}$$
My try : $$100=10^2\\101=(100+1)=(10^2+1)$$
So :
$$100^{101}=10^{2(101)}\\101^{100}=(10^2+1)^{100}=10^{2(100)}+N$$
Now what ?
algebra-precalculusapproximationinequalityupper-lower-bounds
Which of the following numbers is greater? Without using a calculator and logarithm.
$$100^{101} , 101^{100}$$
My try : $$100=10^2\\101=(100+1)=(10^2+1)$$
So :
$$100^{101}=10^{2(101)}\\101^{100}=(10^2+1)^{100}=10^{2(100)}+N$$
Now what ?
Best Answer
You want to determine if $\left(\frac{101}{100}\right)^{100}\geq 100$. But we know that $ \left(1+\frac{1}{n}\right)^n$ is always less than $e$.