Consider the following statements:Statement 1 | vedclass

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(A) For Statement $1$:$y = \log_{10} x + \log_{e} x$Using the change of base formula,$\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we have:$y = \frac{\

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Statement $1$ is true,Statement $2$ is false.

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(A) For Statement $1$:$y = \log_{10} x + \log_{e} x$Using the change of base formula,$\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we have:$y = \frac{\log_{e} x}{\log_{e} 10} + \log_{e} x$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{x \log_{e} 10} + \frac{1}{x}$Since $\frac{1}{\log_{e} 10} = \log_{10} e$,we get:$\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.Thus,Statement $1$ is true.For Statement $2$:$\frac{d}{dx}(\log_{10} x) = \frac{d}{dx} \left( \frac{\log_{e} x}{\log_{e} 10} \right) = \frac{1}{x \log_{e} 10}$.The statement claims it is $\frac{\log x}{\log 10}$,which is incorrect.Similarly,$\frac{d}{dx}(\log_{e} x) = \frac{1}{x}$.The statement claims it is $\frac{\log x}{\log e}$,which is incorrect.Thus,Statement $2$ is false.

Consider the following statements:
Statement $1$: If $y = \log_{10} x + \log_{e} x$,then $\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.
Statement $2$: $\frac{d}{dx}(\log_{10} x) = \frac{\log x}{\log 10}$ and $\frac{d}{dx}(\log_{e} x) = \frac{\log x}{\log e}$.

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Consider the following statements:Statement $1$: If $y = \log_{10} x + \log_{e} x$,then $\frac{dy}{dx} = \frac{\log_{10} e}{x} + \frac{1}{x}$.Statement $2$: $\frac{d}{dx}(\log_{10} x) = \frac{\log x}{\log 10}$ and $\frac{d}{dx}(\log_{e} x) = \frac{\log x}{\log e}$.