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(B) Given: $\cosh (x-\log 3)=\sinh x$Using the definitions $\cosh 	heta = \frac{e^{	heta}+e^{-	heta}}{2}$ and $\sinh 	heta = \frac{e^{	heta}-e^{-

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$\frac{1}{2} \log 6$

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(B) Given: $\cosh (x-\log 3)=\sinh x$Using the definitions $\cosh 	heta = \frac{e^{	heta}+e^{-	heta}}{2}$ and $\sinh 	heta = \frac{e^{	heta}-e^{-	heta}}{2}$:$\frac{e^{x-\log 3}+e^{-(x-\log 3)}}{2} = \frac{e^x-e^{-x}}{2}$$e^x \cdot e^{-\log 3} + e^{-x} \cdot e^{\log 3} = e^x - e^{-x}$Since $e^{-\log 3} = \frac{1}{3}$ and $e^{\log 3} = 3$:$\frac{1}{3} e^x + 3 e^{-x} = e^x - e^{-x}$$3 e^{-x} + e^{-x} = e^x - \frac{1}{3} e^x$$4 e^{-x} = \frac{2}{3} e^x$$e^{2x} = 4 \cdot \frac{3}{2} = 6$$2x = \log 6$$x = \frac{1}{2} \log 6$

If $\cosh (x-\log 3)=\sinh x$,then $x=$

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