If R35 000 accumulates to R48 320 at a continuous compounded rate of 8,6% per year, then the term under consideration is .... years.
[1] 6,23
[2] 2,77
[3] 4,43
[4] 3,91
[5] 3,75
"S=Pe^{ct}"
"48320=35000e^{0.086t}"
"\\frac{48320}{35000}=e^{0.086t}"
"ln(48320\/35000)=lne^{0.086t}=0.086t\\space lne"
"\\frac{ln(48320\/35000)}{0.086}=t"
"t=3,74997"
So, the term under consideration is "3,75"
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