If y=2^{ax} and left(frac{dy}{dx}right)_{x=1} | vedclass

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(B) Given $y=2^{ax}$.Applying the derivative formula $\frac{d}{dx}(b^{f(x)}) = b^{f(x)} \cdot \ln(b) \cdot f'(x)$,we get:$\frac{dy}{dx} = 2^{ax} \cdot

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$2$

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(B) Given $y=2^{ax}$.Applying the derivative formula $\frac{d}{dx}(b^{f(x)}) = b^{f(x)} \cdot \ln(b) \cdot f'(x)$,we get:$\frac{dy}{dx} = 2^{ax} \cdot \ln(2) \cdot a$.At $x=1$,the derivative is $\left(\frac{dy}{dx}\right)_{x=1} = 2^a \cdot a \cdot \ln(2)$.Given that $\left(\frac{dy}{dx}\right)_{x=1} = \log 256$,and knowing $\log 256 = \log(2^8) = 8 \log 2$,we equate the two expressions:$2^a \cdot a \cdot \ln(2) = 8 \ln(2)$.Dividing both sides by $\ln(2)$,we get:$a \cdot 2^a = 8$.By inspection,if $a=2$,then $2 \cdot 2^2 = 2 \cdot 4 = 8$.Thus,$a=2$.

If $y=2^{ax}$ and $\left(\frac{dy}{dx}\right)_{x=1}=\log 256$,then $a=$

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