Question

the eye size of many vertebrates is related to the body mass by the logarithmic equation log(E) = log(10.61) + 0.1964log(m) where E is the eye axial length in millimeters, and m is the body mass in kilograms. Predict the mass of a vertebrate with an eye axial length of 22mm

Answer 1

You want to find the mass given the formula \[\log(E) = \log(10.61) + 0.1964\log(m)\]First thing to do is rearrange that formula to give you \(m\) in terms of everything else. Can you do that?

Answer 2

Not very good at this unit, so no unfortunately

Answer 3

So E is the eye axial length (in mm)
m is body mass (in kg)
it wants to find m when E=22 mm

Answer 4

Insert 22 for E and solve for m

Answer 5

show me as far as you can get
and we will work from there

Answer 6

sorry, I must have made a mistake in my arithmetic, now it looks more reasonable.

Answer 7

just double checked, and that is the formula

if it helps this is a multiple choice question my options are

2.37

444.40

1.15

40.98

Answer 8

don't be scared by the logarithms. this is just algebra.

\[\log(E) = \log(10.61) + 0.1964\log(m)\]we want the term with \(m\) in it alone on one side. what would you do as a first step?

Answer 9

How about we subtract \(\log(10.61)\) from both sides? what do you get if you do that?

Answer 10

log(E) - log(10.61) = 0.1964log(m)

Answer 11

good. now if we want to solve that for \(\log(m)\) what do we do next?

Answer 12

divide both sides by 0.1964 ?

Answer 13

yep

Answer 14

if you want to go ahead and replace E with 22 you can
an approximate the left hand side

Answer 15

you will get something in the form
\[y \approx \log(m) \implies 10^{y} \approx m\]

Answer 16

alright so [log(22) - log(10.61)]/0.1974 = log(m)

Answer 17

cool stuff
and you can enter that stuff on the left hand side into a calculator
i can check your approximation

Answer 18

3.71306

Answer 19

m = 40.9791

Answer 20

looks like you use natural log instead

not a problem we will just have to fix our base for the equivalent exponential form

Answer 21

oh it looks like you did that

Answer 22

Looks good to me!

Answer 23

\[y \approx \log_e(m) \implies e^{y} \approx m\]
great job