A new smartphone can be purchased for $840. The phone’s value has a half-life of 23 months and can be modelled by the function
𝑉 = 840(0.5) 𝑡/23
Algebraically determine how long, to the nearest tenth of a month, it takes the smartphone to be worth $601
"V=840(0.5)^\\frac {t} {23}"
Given "V=601"
"601=840(0.5)^\\frac {t} {23}"
"\\frac {601}{840}=(0.5)^\\frac {t} {23}"
"Log(\\frac {601} {840}) =\\frac {t} {23} Log 0.5"
"-0.145404814=\\frac {t} {23} \u00d7-0.301029996"
"t=\\frac {23\u00d70.145404814}{0.301029996}"
"t=11.1095"
"t=11.11months"
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