Hmmm. 81 - 43 is 38 db.
intentity = antilog (db level/10)
= antilog 3.8= 10^3.8
that is almost 7000 times.
A 60 year old person has a threshold of hearing of 81.0 dB for a sound with frequency f=10,000 Hz. By what factor must the intensity of a sound wave of that frequency, audible to a typical young adult, (sound level=43.0 dB) be increased so that it is heard by the older person.
Thanks for any and all help!
2 answers
Thank You so much Bob Pursley!!