The rate of chirping of the snowy tree cricket ( Oecanthus fultoni Walker) varie

Question

The rate of chirping of the snowy tree cricket (Oecanthus fultoni Walker) varies with temperature in a predictable way. A linear relationship provides a good match to the chirp rate, but an even more accurate relationship is the following:
N=(5.63×1010)e?(6290K)/T
In this expression, N is the number of chirps in 13.0 s and T is the temperature in kelvins.

If a cricket is observed to chirp 177 times in 60.0 s, what is the temperature in degrees Fahrenheit?

I KEEP ON GETTING -5.78 DEGREES FAHRENHEIT....which is wrong...what am I doing wrong? My work:

Explanation / Answer

N=(5.63×1010)e(6290K)/T

N=(177/60)*13= 38.35

38.35=(5.63×1010)e(6290K)/T

ln38.35=ln[(5.63×1010)e(6290K)/T]

ln38.35=ln(5.63×1010) +lne(6290K)/T

ln38.35=ln(5.63×1010) +((6290K)/T)lne

ln38.35=ln(5.63)+ln(1010) (6290K)/T

ln38.35=ln(5.63)+10ln(10) (6290K)/T

(6290K)/T =-ln38.35+ln(5.63)+10ln(10)

(6290K)/T =

T=6290K/21.10721

T=298K

temperature in degrees C =298-273.15 =24.85oC

temperature in degrees F =(24.85*(9/5))+32=76.73oF

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