Scott92
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- Joined
- Jun 11, 2022
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- 41
I have the following question but two ways to solve it not sure which if either is correct
Differentiate each of the following voltage functions with respect to time and hence determine the ‘rate of change’ for each of the functions when time (t) is 5 seconds v=loge(2t)
loge(2×5)
loge(2×5)=loge(2)+loge+(5) Use log rule loge(x)=In(x)
In(2)+loge(5)=In(2)+In(5)=2.302585093 vs
0r simply 1/t 1/5=0.2vs
Differentiate each of the following voltage functions with respect to time and hence determine the ‘rate of change’ for each of the functions when time (t) is 5 seconds v=loge(2t)
loge(2×5)
loge(2×5)=loge(2)+loge+(5) Use log rule loge(x)=In(x)
In(2)+loge(5)=In(2)+In(5)=2.302585093 vs
0r simply 1/t 1/5=0.2vs