Question

Help please

Y=Ae^bx

Please solve 5 (b)

Thanks

Answer 1

Answer:

Step-by-step explanation:

Assuming you have solved a part A=4,B= -1/10

Now Take a Derivative wrt x

dy/dx=gradient= -1 =ABe^(bx)

putting values of A & B

x=1/4*ln(5/2)

again put this value in the equation to get y

Answer 2

$HEYA \: \\ \\ y = ae {}^{bx} \\ \\ a \: part)\\ \\ (0 \: \: 4) \: \: \: x = 0 \: \: \: and \: \: \: y = 4 \\ put \: these \: two \: \: values \: \: of \: x \: and \: y \: in \: above \: \\ equation \: we \: have \: \\ \\ 4 = ae {}^{b \times 0} \\ \\ a = 4 \\ \\ (10 \: \: \: 4 \div e) \: \: \: x = 10 \: \: \: and \: \: \: y = 4 \div e \\ put \: these \: values \: of \: x \: and \: y \: in \\ \: above \: equation \: we \: have \\ \\ (4 \div e) = ae {}^{10b} \\ \\ (4 \div e) = 4 \times e {}^{10b} \\ \\ (1 \div e) = e {}^{10b} \\ \\ e {}^{ - 1} = e {}^{10b} \\ take \: ln \: on \: both \: sides \: we \: have \\ \\ ln(e {}^{ - 1} ) = ln(e {}^{10b} ) \\ \\ - 1 = 10b \\ \\ b = - (1 \div 10) \\ \\ \\ b \: part \: ) \\ \\ y = ae {}^{bx} \\ \\ take \: ln \: on \: both \: sides \: we \: have \\ \\ ln(y) = ln(a) + bx \: ....Equation \: i \: \: \\ \\ Differentiate \: both \: sides \: with \: \\ respect \: to \: x \: we \: have \\ \\ (1 \div y)dy \div dx = b \\ \\ dy \div dx = yb \: \: \: \: \: \: \: (dy \div dx) = - 1 \\ \\ - 1 = yb \\ \\ - 1 = - (1 \div 10)y \\ \\ y = 10 \\ \\ now \: put \: value \: of \: y \: in \: equation \: i \: we \: have \\ \\ ln(10) = ln(4) + bx \\ \\ bx = ln(10) - ln(4) \\ \\ bx = ln(10 \div 4) \\ becoz \: \: ln(m) - ln(n) = ln(m \div n) \\ \\ bx = ln(2.5) \\ \\ x ( - 1 \div 10) = ln(2.5) \\ \\ x = - 10 ln(2.5) \\ \\ so \: the \: exact \: coordinates \: are \: \: ( - 10 ln(2.5) \: \: 10)$