Please solve this
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Answers
Answer:
a)will b) are I think it was helpful
Answer:
I am in class 10th dear why are you asking dear do u want to say something
Let ,
\small \bf y = log_{x}(a) + log_{a}(x) + \bigg \{log_{x}(a) . log_{a}(x) \bigg \}y=logx(a)+loga(x)+{logx(a).loga(x)}
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▪To Calculate :-
\bf \pink{ \dfrac{dy}{dx} }dxdy
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▪Formulae Used :-
\begin{gathered} \maltese \: \: \: \bf log_{n}(m) = \dfrac{ log_{e}(m) }{ log_{e}(n) } \\ \\ \maltese \: \: \: \bf \frac{d}{dx} \: log \: x = \frac{1}{x} \\ \\ \maltese \: \: \: \bf \frac{d}{dx} \: \{ f(x) { \}}^{n} = n. \{f(x) \} {}^{n - 1} . \frac{d}{dx} \: f(x) \end{gathered}✠logn(m)=loge(n)loge(m)✠dxdlogx=x1✠dxd{f(x)}n=n.{f(x)}n−1.dxdf(x)
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▪Solution :-
\begin{gathered} \sf\text{ y }= log_{x}(a) + log_{a}(x) + \big\{log_{x}(a) . log_{a}(x) \} \\ \\ \sf= log_{x}(a) + log_{a}(x) + \bigg \{ \dfrac{ log_{e}(a) }{ log_{e}(x) } \times \frac{ log_{e}(x) }{ log_{e}(a) } \bigg \}\end{gathered} y =logx(a)+loga(x)+{logx(a).loga(x)}=logx(a)+loga(x)+{loge(x)loge(a)×loge(a)loge(x)}
\begin{gathered} \implies \sf y = log_{x}(a) + log_{a}(x) + 1\\ \\ \implies \bf y = \frac{ log_{e}(a) }{ log_{e}(x) } + \frac{ log_{e}(x) }{ log_{e}(a) } + 1\end{gathered}⟹y=logx(a)+loga(x)+1⟹y=loge(x)loge(a)+loge(a)loge(x)+1
Let ,
\begin{gathered} \sf A = \frac{ log_{e}(a) }{ log_{e}(e) } \\ \\ \sf B = \frac{ log_{e}(x) }{ log_{e}(a) }\end{gathered}A=loge(e)loge(a)B=loge(a)loge(x)
So ,
\begin{gathered} \sf \dfrac{dy}{dx} = \dfrac{dA}{dx} + \dfrac{dB}{dx} + \dfrac{d1}{dx} \\ \\ \implies \sf \dfrac{dy}{dx} = \dfrac{dA}{dx} + \dfrac{dB}{dx} + 0 \\ \\ \implies\bf \dfrac{dy}{dx} = \dfrac{dA}{dx} + \dfrac{dB}{dx}\:\:\:\:.\:.\:.\:\{i\}\end{gathered}dxdy=dxdA+dxdB+dxd1⟹dxdy=dxdA+dxdB+0⟹dxdy=dxdA+dxdB...{i}
We have ,
\begin{gathered} \sf A = \frac{ log_{e}(a)}{ log_{e} (x)} \\ \\ \sf A = log_{e}(a) \: . \: \frac{1}{ log_{e}(x) } \end{gathered}