Not sure if I’m correct

Dr.Peterson

Elite Member
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

Senior Member
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

New member
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

New member
Or is the slope just equal to log(5)/2 which would be equal to 0.35?

Otis

Senior Member
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

Dr.Peterson

Elite Member
Or is the slope just equal to log(5)/2 which would be equal to 0.35?
That looks more like it ...

• jazz2112

jazz2112

New member
Thankyou for the help! 