Not sure if I’m correct

jazz2112

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Mar 21, 2019
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This is my working out and what I think the answer is, but I don’t think I have it right, could someone please help :)
12240
 

Dr.Peterson

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Nov 12, 2017
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You've given the intercept and slope for the line v = 3.8 + 5/2 t. That is irrelevant to the problem.

What are the intercept and slope for the line you identified, v = log(3.8) + log(5/2) t?

Then go back and fix that equation, too! It is not quite true that log(5^{t/2}) = t log(5/2). Do you see where the 2 should really be? Then you can answer the question using the correct line.
 

Otis

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Apr 22, 2015
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Hello. You correctly wrote the following property, but you made a mistake when you applied it.

log(an) = n∙log(a)

We see on the left-hand side (above) that n is the exponent.

log(5t/2)

That is, the base is 5 and the exponent is t/2.

In the symbolic expression log(an), we have a=5 and n=t/2.

Therefore, what is n∙log(a) ?

😎
 

jazz2112

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Mar 21, 2019
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Okay so is it supposed to read
V = log(3.8) + t/2• log(5)
If so does that then mean the slope is 5? Or how do I move the 2 from there?
Thankyou!!
 

jazz2112

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Mar 21, 2019
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Or is the slope just equal to log(5)/2 which would be equal to 0.35?
 

Otis

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Or is the slope ... log(5)/2 ...
Yes. In the form:

y = m∙x + b

we see that slope m is the Real number multiplying independent variable x.

In the form:

V = log(5)/2∙t + log(3.8)

we see that expression log(5)/2 is the Real number multiplying independent variable t.

😎
 

jazz2112

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Mar 21, 2019
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Thankyou for the help! 😄
 
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