After x weeks of prcatice, a student can type f(x)=80(1-e^-0.3x)words a minute. How sooon will the student be able to type 60 words per minute? (Round to the nearest tenth)
60=80(1-e-.3x)
3/5=1-e^-.3x)
2/5=e^-.3x
ln (2/5)=-3.x
Surely you can solve it from here. This is hardly calculus.
To find out how soon the student will be able to type 60 words per minute, we need to determine the value of "x" when f(x) equals 60.
The given function is f(x) = 80(1 - e^(-0.3x)). We want to find when f(x) becomes 60.
Setting f(x) equal to 60, we have:
60 = 80(1 - e^(-0.3x))
Now, let's solve for x step by step.
Divide both sides of the equation by 80:
60/80 = 1 - e^(-0.3x)
Rearrange the equation:
3/4 = 1 - e^(-0.3x)
Subtract 1 from both sides of the equation:
-1/4 = -e^(-0.3x)
Multiply both sides of the equation by -1 to isolate the exponential term:
1/4 = e^(-0.3x)
To eliminate the exponential, take the natural logarithm (ln) of both sides:
ln(1/4) = ln(e^(-0.3x))
Using the property of logarithms (ln(a^b) = b*ln(a)), we have:
ln(1/4) = -0.3x * ln(e)
Recall that ln(e) equals 1:
ln(1/4) = -0.3x * 1
Now divide both sides of the equation by -0.3 to isolate "x":
ln(1/4) / -0.3 = x
Using a calculator, evaluate the left side:
x ≈ -2.38021
Since "x" represents the number of weeks, we cannot have a negative number of weeks. Therefore, we disregard the negative sign and take the absolute value:
x ≈ 2.38021
Rounding to the nearest tenth, we have:
x ≈ 2.4
Therefore, the student will be able to type 60 words per minute in approximately 2.4 weeks.