Question 931525
The way I understand the problem, the Main Range Trail ascends eastbound out of Yakker's Lake according to
{{{H(x)= 0.75+(2/5)log((x+1))}}} ,
where {{{x}}} is elapsed distance in miles eastbound from the lake,
{{{H(x)}}} is miles above the elevation of the lake,
and {{{log((x+1))}}} is {{{log(10,(x+1))}}} , the common logarithm (base 10 logarithm) .
I also understand that Mathematics Enthusiast magazine reports that the slope of the graph for a function of the form
{{{y=C+k*log(b,(x+1))}}} is {{{(k/ln(b))*(1/(x+1))}}} for any positive value of x.
 
{{{H(x)= 0.75+(2/5)log(10,(x+1))}}} is a function of the form {{{y=C+k*log(b,(x+1))}}} ,
with {{{C=0.75}}} , {{{b=10}}} , and {{{k=2/5=0.4}}} ,
so the slope of its graph is given by
{{{(0.4/ln(10))*(1/(x+1))}}} .
So, the slope/grade of the trail east of Yakker's Lake is represented by the function
{{{grade=0.4/(ln(10)*(x+1))}}} .
 
Fred begin to think he is hiking on the flat when {{{grade=0.03}}} , so our equation is
{{{0.03=0.4/(ln(10)*(x+1))}}} .
Solving:
{{{0.03=0.4/(ln(10)*(x+1))}}}
{{{0.03*(x+1)=0.4/ln(10)}}}
{{{x+1=0.4/(ln(10)*0.03)}}}
{{{x=0.4/(ln(10)*0.03)-1}}}
Using the approximate value {{{ln(10)=2.3026}}}
(or using however many digits my calculator can handle),
{{{x=4.79}}} (rounded).
So, Fred starts thinking the trail is flat when he is 4.8 miles east of the lake.