Logarithm Word Problem

[thunc14]

Four snail friends are out racing. They are able to go fast at the beginning, but they become tired and their distances from the starting point are modeled by the following equations:
k(x) = (1/3)log(x)+3
g(x) = log(x) + log(x-1)
h(x) = log(2x)-log(x+1)
f(x) = 2log(x)

When will the slowest one get to 5 miles, if x represents time in minutes?
a) about 5.27 hours
b) about 5.28 hours
c) about 5 hours
d) about 1.90 years
e) Answer is not there

I plugged all of the equations into desmos, and h(x) = log(2x) - log (x+1) is the slower snail. When I set that equal to 5 in wolfram alpha, I got x= -(e^-5/(e^5-2)), which doesn't make sense in this case. Am I doing something wrong, or is there no real solution, meaning answer e

 

[Romsek]

\(\displaystyle \lim \limits_{x\to\infty} h(x) = \ln(2) < 5\)

The 3rd snail is the slowest as you correctly determined and the above limit shows it never reaches the 5 mile point.

Answer (e) is correct.

 

[Steven G]

\(\displaystyle \lim \limits_{x\to\infty} h(x) = \ln(2) < 5\)

The 3rd snail is the slowest as you correctly determined and the above limit shows it never reaches the 5 mile point.

Answer (e) is correct.

I think that you need to be more careful than that. Maybe the (continuous) function goes above 5 (or reaches just 5) and then approaches ln(2) for large values of x.

 

[Deleted member 4993]

I think that you need to be more careful than that. Maybe the (continuous) function goes above 5 (or reaches just 5) and then approaches ln(2) for large values of x.

You are correct in "warning" - in general.

However, these functions do not go through any local maxima/minima - so there is no danger of "whip-lash".