Using some combination of algebra and/or with the aid of a scientific calculator (with a log button),
determine the minimum integer x that satisfies the following:
$\displaystyle 2^{x + 1} \ > \ 1.9^x \ + \ 1.9^{x + 1} \ + \ 1.9^{x + 2}$
Using some combination of algebra and/or with the aid of a scientific calculator (with a log button),
determine the minimum integer x that satisfies the following:
$\displaystyle 2^{x + 1} \ > \ 1.9^x \ + \ 1.9^{x + 1} \ + \ 1.9^{x + 2}$
$\displaystyle 1.9^x+ 1.9^{x+1}+ 1.9^{x+2}= 1.9^x+ (1.9)1.9^x+ (1.9^2)(1.9^x)= (1+ 1.9+ 1.9^2)1.9^x= 6.51(1.9^x)$
so that can be written as 2(2^x)> 0.51(1.9^x). Since those numbers are all positive, that is the same as $\displaystyle \frac{2^x}{1.9^x}= 1.05^x> 2.55$. And since logarithm is an increasing function, log(1.05^x)= x log(1.05)> log(2.55).
Making the reasonable assumption that we can expand $\displaystyle 1.9^x$ as $\displaystyle (2 - 0.1)^x$ I'm going to do a Taylor series to first order:
$\displaystyle (2 - 0.1)^x \approx 2^{x - 1} - 2 \cdot 1.9 ~ ln(x) 2^{x - 1} = 2^{x - 1} (1 - 2 \cdot 1.9 x)$
So
$\displaystyle 1.9^x + 1.9^{x+1} + 1.9^{x + 2} \approx 2^{x-1} (1 - 2 \cdot 1.9 x) + 2^x (1 - 2 \cdot 1.9 (x + 1) ) + 2^{x + 1} (1 - 2 \cdot 1.9 (x + 2) )$
So we are looking for the samllest integer x such that
$\displaystyle 1.9^x + 1.9^{x+1} + 1.9^{x + 2} \approx 2^{x-1} (1 - 2 \cdot 1.9 x) + 2^x (1 - 2 \cdot 1.9 (x + 1) ) + 2^{x + 1} (1 - 2 \cdot 1.9 (x + 2) ) = 2^{x + 1}$
I have the graph attached below.
So we know that there is, in fact a lowest x such that the inequality is true.
Okay, I admit it, I have to use W|A to solve this numerically. I did this and found that the two sides are equal when $\displaystyle x \approx -1.31579$. So the smallest integer which satisfies the condition will be x = -1.
-Dan
But is this true from HallsofIvy?
$\displaystyle x \ > \ \dfrac{log(2.55)}{log(1.05)} \ = \ 19.1...$
The next integer is 20.
Does 20 work?
$\displaystyle 2^{20 + 1} \ > \ 1.9^{20} \ + \ 1.9^{20 + 1} \ + \ 1.9^{20 + 2} \ $?
$\displaystyle 2^{21} \ > \ 1.9^{20} \ + \ 1.9^{21} \ + \ 1.9^{22} \ $?
$\displaystyle 2,097, 152 \ > \ 2,447,107.27... \ \ \ \ \ \ \ $
This is not true. It must be a different integer than 20.
$\displaystyle \dfrac{2^x}{1.9^x} \ > \ 3.255$
$\displaystyle \bigg(\dfrac{2}{1.9}\bigg)^x \ > \ 3.255$
$\displaystyle (x)log\bigg(\dfrac{2}{1.9}\bigg) \ > \ log(3.255)$
$\displaystyle x \ > \ \dfrac{log(3.255)}{log(\tfrac{2}{1.9})}$
$\displaystyle x \ > \ 23.0087...$
$\displaystyle x \ = \ 24 \ \ $works.