Exponent + Inequality Problem

[Marghe]

Hello,

I can not post the source of the problem for copyright reasons. Therefore, I write a similar problem with different values.

What is the largest integer m such that 1/3^m > .08?

A) 5
B) 6
C) 7
D) 10
E) 51


If I prove, 1/3^7 = .0004572<.08
I need a smaller denominator, so I prove 1/3^6= .0013717>.08
Therefore, m is 6.

Could you please show me a different way that involves less computation?

Thanks.

[williamfmarshall]

Hi there, I'm not sure the set of answers provided has the correct choice. The question seems to be what is the largest exponent m such that the result of (1/3)^m will still be larger ~1/12 (.08). If we start with m=2 the result is (1/3 x 1/3=1/9) which is still larger than 1/12.

The next positive integer 3 would be (1/3 x 1/3 x 1/3=1/27) which is less than 1/12 and doesn't satisfy the inequality. So your answer appears to be 2 for the inequality as it is written right now.

Hope this helps.

[RonPurewal]

The previous poster is correct; the answer to this problem as currently written is 2.

One way to do less computation is to flip both sides over (take reciprocals).
1/3^n > 0.08

3^n < 1/0.08
(note: reciprocals of bigger numbers are smaller, and vice versa; therefore, flip the inequality)

The value of 1/0.08 is equivalent to 100/8, or 12.5. Therefore, we just want the biggest n such that 3^n is less than 12.5.

[Marghe]

Thanks!

I now understand the original solution that unfortunately does not provide the explanation about the flip step; I was not so prompt to assume the movement.

[RonPurewal]

Thanks!

I now understand the original solution that unfortunately does not provide the explanation about the flip step; I was not so prompt to assume the movement.

I don't understand what this means.