What is the tangent of f(x)=ln(x^3),\at x=e^5 ?

The tangent of f(x)=ln(x^3),\at x=e^5 is y= 3/(e^5)x+12

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tangent of f(x)=ln(x^3),\at x=e^5

tangent of

f (x) = ln (x^{3}), at x = e^{5}

y = \frac{3}{e ^{5}} x + 12

Solution steps

Find the tangent point:

(e^{5}, 15)

Find the slope of

f (x) = ln (x^{3}) : \frac{df ( x )}{d x} = \frac{3}{x}

Compute the slope of

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

x = e^{5} : m = \frac{3}{e ^{5}}

Find the line with slope m=

\frac{3}{e ^{5}}

and passing through

(e^{5}, 15) : y = \frac{3}{e ^{5}} x + 12

y = \frac{3}{e ^{5}} x + 12

f (x) = (x^{3} + 6 x)^{4}, at x = 2

\int_{0}^{2} \frac{2 x}{( x ^{2} + 3 ) ^{3}} d x

x^{2} + x

sin (\frac{π x}{3})

\int \frac{( 4 + x + x ^{2} )}{x} d x

What is the tangent of f(x)=ln(x^3),\at x=e^5 ?
The tangent of f(x)=ln(x^3),\at x=e^5 is y= 3/(e^5)x+12