Question

How to prove this without mathematical induction?
$\implies 2^{x}\leq2\times(x!)$

Answer 1

Answer:

5 IST 5 =10

Step-by-step explanation:

Answer 2

$\large\underline{\text{Solution}}$

Consider the two logarithms.

$\implies\log2^{x}$

$\implies\log2x!$

$\red{\bigstar}$The properties of logarithms.

$\implies\log a+\log b=\log ab$

$\implies\log m^{n}=n\log m$

Then

$\implies\log2^{x}=x\log2$

$\implies\log2x!=\log2+(\log1+\log2+\cdots+\log x)$

We divide both by $\log2$.

$\implies\dfrac{x\log2}{\log2}=x$

$\implies\dfrac{\log2x!}{\log2}=1+\dfrac{\log1+\log2+\cdots+\log x}{\log2}$

Then let's check the second equation.

$\implies1+\dfrac{\log1+\log2+\cdots+\log x}{\log2}=1+\dfrac{\log1}{\log2}+\dfrac{\log2}{\log2}+\cdots+\dfrac{\log x}{\log2}$

$\red{\bigstar}$We spot the series of logarithms.

$\implies a_{n}=\dfrac{\log n}{\log2}$

$\red{\bigstar}$Each of the terms compares to 0 and 1.

$a_{n}=\begin{cases} & \text{for }n=1, a_{n}=0 \\ & \text{for }n=2, a_{n}=1 \\ & \text{for }n\geq3, a_{n}>1\end{cases}$

Thus

$\implies1+\dfrac{\log1}{\log2}+\dfrac{\log2}{\log2}+\cdots+\dfrac{\log x}{\log2}\geq1+0+1+1+\cdots+1=x$

Hence

$\implies\dfrac{\log2x!}{\log2}\geq x$

We multiply the inequality by $\log2$.

$\implies\log2x!\geq x\log2$

$\implies x\log2\leq\log2x!$

$\implies \log2^{x}\leq\log2x!$

Thereby

$\implies 2^{x}\leq2x!\text{ (Equals when }x=1\text{ or }x=2\text{)}$

Hence proven.

$\large\underline{\text{Answers of different methods}}$

(Mathematical induction method)

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