Thread: minimum integer solution for an inequality

1. minimum integer solution for an inequality

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:

$\displaystyle 2^{x + 1} \ > \ 1.9^x \ + \ 1.9^{x + 1} \ + \ 1.9^{x + 2}$

2. Re: minimum integer solution for an inequality

$\displaystyle 1.9^x+ 1.9^{x+1}+ 1.9^{x+2}= 1.9^x+ (1.9)1.9^x+ (1.9^2)(1.9^x)= (1+ 1.9+ 1.9^2)1.9^x= 6.51(1.9^x)$
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).

3. Re: minimum integer solution for an inequality

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

4. Re: minimum integer solution for an inequality

Originally Posted by HallsofIvy
$\displaystyle 1.9^x+ 1.9^{x+1}+ 1.9^{x+2}= 1.9^x+ (1.9)1.9^x+ (1.9^2)(1.9^x)= (1+ 1.9+ 1.9^2)1.9^x= 6.51(1.9^x)$
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).
Okay, much simpler than my idea.

@everyone: So where did I go wrong? Should I have done more terms in the Taylor series? I should have come close at least.

-Dan

5. Re: minimum integer solution for an inequality

Originally Posted by HallsofIvy
..., log(1.05^x)= x log(1.05)> log(2.55).
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.

6. Re: minimum integer solution for an inequality

Originally Posted by greg1313
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 \begin{align*}2^{x + 1} &> 1.9^x \ + 1.9^{x + 1} + 1.9^{x + 2}\\2\cdot 2^x&>1.9^x(1+1.9+1.9^2)\\2\cdot 2^x&>(6.51)1.9^x\\2^x&>3.255(1.9^x) \end{align*} SEE HERE1, SEE HERE2 ,

SEE HERE3 So it seems that $x=24$ works.

7. Re: minimum integer solution for an inequality

Originally Posted by Plato
\displaystyle \begin{align*}2^{x + 1} &> 1.9^x \ + 1.9^{x + 1} + 1.9^{x + 2}\\2\cdot 2^x&>1.9^x(1+1.9+1.9^2)\\2\cdot 2^x&>(6.51)1.9^x\\2^x&>3.255(1.9^x) \end{align*}
$\displaystyle \dfrac{2^x}{1.9^x} \ > \ 3.255$

$\displaystyle \bigg(\dfrac{2}{1.9}\bigg)^x \ > \ 3.255$

$\displaystyle (x)log\bigg(\dfrac{2}{1.9}\bigg) \ > \ log(3.255)$

$\displaystyle x \ > \ \dfrac{log(3.255)}{log(\tfrac{2}{1.9})}$

$\displaystyle x \ > \ 23.0087...$

$\displaystyle x \ = \ 24 \ \$works.
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