Solved Examples related to the Value of e
Let’s see some solved examples on Value of e.
Solved Example 1: Find the value of r if: \(re^{-3} = e^{-2}(lne^{-2})^2\)
Solution:
\(re^{-3} = e^{-2}(lne^{-2})^2\)
From the properties of logarithm we know that, \(log_{10}a^n = nlog_{10}a\)
\(re^{-3} = e^{-2}(-2lne)^2\)
lne = 1, as the base is same.
\(re^{-3} = e^{-2}(-2)^2\)
\(re^{-3} = e^{-2}4\)
\(r = e^{-2 + 3}4\)
\(r = e\times4\)
\(r = 10.87\).
Solved Example 2: Find the derivative of \(y=e^x\)
Solution:
\(y=e^x\)
Let’s take log to base e on both sides
\(lny = lne^x\)
From the properties of logarithm we know that, \(log_{10}a^n = nlog_{10}a\)
\(lny = xlne\)
lne = 1, as the base is same.
lny = x
Differentiating on both the sides,
\({1\over{y}}{dy\over{dx}} = {d\over{dx}}{x}\) …….. \([{d\over{dx}}logx = {1\over{x}}]\)
Thus, \({dy\over{dx}} = y(1) = y …….. [{dx\over{dx}}=1]\)
\({dy\over{dx}} = e^x\)
Solved Example 3: Evaluate \(f(x) = e^x\) to the nearest thousandth for each value of x below.
Solution:
\(f(x) = e^x\)
x = 2
\(e^2 = 7.389\)
x = ½
\(e^{1\over2} = 1.649\)
x = -1
\(e^-1 = 0.368\)
x = 6
\(e^6 = 403.429\)
\( x = {1\over{3}} \)
\(e^{1\over{3}} = 1.396\)
x = -2
\(e^{-2} = 0.135\)
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