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Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
sarahl wrote:
On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 165 is how many times an intensity corresponding to a magnitude of 125?


(A) 40
(B) 100
(C) 400
(D) 1,000
(E) 10,000



Putting this into words, we can express it this way:

Let \(x\) by the factor of increase.

So if magnitude goes from 0 to 10, then intensity increases from 0 to 10.
If magnitude goes from 10 to 20, then intensity increases tenfold, so it goes from 10 to 100.
Magnitude increases from 20 to 30, then intensity increases from 100 to 1000.
And so on.

Following this pattern, we can say that:

\(10x => 10^x\), or that \(10x\) corresponds to \(10^x\).

Now we can solve for \(x\) in both magnitude scenarios and find their corresponding intensities:

Magnitude = 165

\(10x = 165\)
\(x = 16.5\)

Which means that:

\(10^x = 10^{16.5}\)


Magnitude = 125

\(10x = 125\)
\(x = 12.5\)

Which means that:

\(10^x = 10^{12.5}\)


Now that we have our intensities, we can compare them:

\(\frac{10^{16.5}}{10^{12.5}} = 10^4 = 10,000\)

So the answer is E
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Re: On a certain scale of intensity, each increment of 10 in mag [#permalink]
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