(C) Given $2 \sinh x = \cosh x$. Using the definitions $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$: $2 \left( \frac{e^x

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(C) Given $2 \sinh x = \cosh x$. Using the definitions $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$: $2 \left( \frac{e^x - e^{-x}}{2} \right) = \frac{e^x + e^{-x}}{2}$ $e^x - e^{-x} = \frac{e^x + e^{-x}}{2}$ $2e^x - 2e^{-x} = e^x + e^{-x}$ $e^x = 3e^{-x}$ $e^{2x} = 3$ Taking the natural logarithm on both sides: $2x = \log_e 3$ $x = \frac{1}{2} \log_e 3$

Class 11 Mathematics Exponential and Logarithmic Series Exponential series

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If $2 \sinh x = \cosh x$,then $x =$

AP EAMCET 2023 All AP EAMCET

A
$\frac{1}{3} \log 2$
B
$2 \log 3$
C
$\frac{1}{2} \log 3$
D
$\log 9$

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Exponential and Logarithmic Series

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