OG 11 #99

maolivie wrote:How would you guys explain this reasoning?

On a scale that measures the intensity of a certain phenomenon, a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?

A) 5
B) 50
C) 10^5
D) 5^10
D) 8^10-3^10

According to the solution, each increase of 1 in the scale creats an intensity increase of a factor of 10? How is that determined?

the darkened part answers ur question .. wen u move from n to n+1 the intensity increases by 10 times .. and n+1 is 1 greater than n on the scale .. post if any more doubts ...

maolivie wrote:How would you guys explain this reasoning?

On a scale that measures the intensity of a certain phenomenon, a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?

A) 5
B) 50
C) 10^5
D) 5^10
D) 8^10-3^10

According to the solution, each increase of 1 in the scale creats an intensity increase of a factor of 10? How is that determined?

Just elaborating on Gabriel's point, this is like the Richter scale (for Earthquakes).

A value of 2 is 10 times the value of 1. 3 is 10 times 2 which is
100 times 1. By induction,

Intensity at level n = 10^n * intensity-at-level-1

So, if you know the intensity at 3 say x

Intensity at 4 = 10 * intensity at 3
Intensity at 5 = 10^2 * intensity at 3
...
Intensity at 8 = 10^5 * intensity at 3

On a scale that measures the intensity of a certain phenomenon, a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?

A) 5
B) 50
C) 10^5
D) 5^10
D) 8^10-3^10

Reading n --- intensity x (assume)
Reading n+1 ---- intensity 10x
Reading n+2 ---- intensity 10(10x) = 100x
Reading 3 ---- intensity x
Reading 4 i.e. n+1 ---- intensity 10x
Reading 8 --- intensity 10^5

Hey, thanks for your replies.

So...

does a reading of 8 (mean 7+1), equate to 10^8, and then a reading of 3 (2+1) equate to 10^3? Then it becomes the difference between the two, so you subtract?

maolivie wrote:Hey, thanks for your replies.

So...

does a reading of 8 (mean 7+1), equate to 10^8, and then a reading of 3 (2+1) equate to 10^3? Then it becomes the difference between the two, so you subtract?

You shouldn't subtract as you are asked to find the "number of times"
one is greater than the other.