Stupid math question

[pitt1166]

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OK, so I'm doing a problem from Kaplan's Q-bank, and the answer they give has me confused.

So they give us Equation 1...

I=I(subscript o)^-bX.

And they tell us that I/I(subscript) ratio is inversely proportional to energy.

The question is, what will be the affect of increasing b by 5 on the energy?

ex. I=I(subscript)^-5bx

They tell me that this decreases the I/I(subscript) ratio by a power of 5 thus increasing the energy. My answer was the opposite... higher I/I(subscript) ratio and thus lower energy. Am I making an extremely stupid math mistake here?

Thanks folks.

 

[TheBoondocks]

OK, so I'm doing a problem from Kaplan's Q-bank, and the answer they give has me confused.

So they give us Equation 1...

I=I(subscript o)^-bX.

And they tell us that I/I(subscript) ratio is inversely proportional to energy.

The question is, what will be the affect of increasing b by 5 on the energy?

ex. I=I(subscript)^-5bx

They tell me that this decreases the I/I(subscript) ratio by a power of 5 thus increasing the energy. My answer was the opposite... higher I/I(subscript) ratio and thus lower energy. Am I making an extremely stupid math mistake here?

Thanks folks.

Based on what you wrote, you're correct. Kaplan Q bank has lots of errors. However, increase generally means addition. Even then, you're still correct. Raising a higher negative power always results in a smaller number. Since we're dividing by a smaller number, the ratio gets bigger. Consequently, energy goes down.

 

[engineeredout]

Looks like you're correct. I = Io^-bx == I = 1/Io^bx. The bigger the b, the smaller the I will be.