Using differentials, find the approximate value of , [tex]\tt\log_{e}(100.1)[/tex] upto three places of decimal.
[tex] \rule{300pt}{0.1pt}[/tex]
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[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that
[tex]\sf \: f(x) = log_{e}(x), \: \: where \: x = 100 \\ \\ [/tex]
So,
[tex]\sf \: f(x + \triangle x) = log_{e}(x + \triangle \: x), \: \: where \: \triangle x = 0.1 \\ \\ [/tex]
Now, using definition of differentials, we have
[tex]\sf \: f(x + \triangle x) = f(x) +f'(x) \triangle x \\ \\ [/tex]
[tex]\sf \: log_{e}(x + \triangle x) = log_{e}(x) + \dfrac{1}{x} \times \triangle x \\ \\ [/tex]
[tex]\sf \: log_{e}(x + \triangle x) = log_{e}(x) + \dfrac{\triangle x}{x} \\ \\ [/tex]
So, on substituting the values, we get
[tex]\sf \: log_{e}(100 + 0.1) = log_{e}(100) + \dfrac{0.1}{100} \\ \\ [/tex]
[tex]\sf \: log_{e}(100.1) = log_{e}( {10}^{2} ) + 0.001 \\ \\ [/tex]
[tex]\sf \: log_{e}(100.1) = 2log_{e}( {10} ) + 0.001 \\ \\ [/tex]
[tex]\sf \: log_{e}(100.1) = 2 \times 2.3026 + 0.001 \\ \\ [/tex]
[tex]\sf \: log_{e}(100.1) = 4.6052 + 0.001 \\ \\ [/tex]
[tex]\bf\implies \:\: log_{e}(100.1) = 4.6062 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]
Step-by-step explanation:
[tex] \large{ \underline{ \bold{Solution - } }}[/tex]
[tex]f(x)= log_e \: where \: \\ x=100
[/tex]
[tex] \bold{f(x+△x)=loge(x+△x),where△x=0.1}[/tex]
[tex] \bold{Now, \: using definition \: of \: \: differentials, we \: have} \\ [/tex]
f (x + △ x)= f (x)+ f (x)△x
[tex]loge(x+△x)=loge(x)+x1×△x[/tex]
So, on substituting the values, we get
[tex]\begin{gathered}\sf \: log_{e}(100 + 0.1) = log_{e}(100) + \dfrac{0.1}{100} \\ \\ \end{gathered}[/tex]
[tex] log_e(100.1)= log_e(102)+0.001 [/tex]
[tex]\begin{gathered}\sf \: log_{e}(100.1) = 2 \times 2.3026 + 0.001 \\ \\ \end{gathered}[/tex]
[tex]loge(100.1)=2×2.3026+0.001[/tex]
[tex] : ⟹ log_e (100.1)=4.6062[/tex]
[tex] \sf{Answer : ∴ Log_e ( 100.1 ) = 4.6062 }[/tex]