What is the tangent of f(x)=sqrt(3)^x,\at x=8 ?

The tangent of f(x)=sqrt(3)^x,\at x=8 is y=(81ln(3))/2 x+81-324ln(3)

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tangent of f(x)=sqrt(3)^x,\at x=8

tangent of

f (x) = 3^{x}, at x = 8

y = \frac{81 ln ( 3 )}{2} x + 81 - 324 ln (3)

Solution steps

Find the tangent point:

(8, 81)

Find the slope of

f (x) = (3)^{x} : \frac{df ( x )}{d x} = \frac{1}{2} ln (3) e^{\frac{l n ( 3 ) x}{2}}

Compute the slope of

f (x) = (3)^{x}

x = 8 : m = \frac{81 ln ( 3 )}{2}

Find the line with slope m=

\frac{81 ln ( 3 )}{2}

and passing through

(8, 81) : y = \frac{81 ln ( 3 )}{2} x + 81 - 324 ln (3)

y = \frac{81 ln ( 3 )}{2} x + 81 - 324 ln (3)

x \to 0 lim (\frac{( x + 1 )}{x})

f (x) = \frac{1}{x ^{5}}, (1, 1)

\int_{0}^{1} 1 + 81 x d x

x \to 3 lim (lo g_{3} (2 x^{3}))

\int x^{2} - 2 x^{4} d x

What is the tangent of f(x)=sqrt(3)^x,\at x=8 ?
The tangent of f(x)=sqrt(3)^x,\at x=8 is y=(81ln(3))/2 x+81-324ln(3)