What is the tangent of y=5^{x-3},(3,1) ?

The tangent of y=5^{x-3},(3,1) is y=ln(5)x+1-ln(5)*3

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tangent of y=5^{x-3},(3,1)

tangent of

y = 5^{x - 3}, (3, 1)

y = ln (5) x + 1 - ln (5) \cdot 3

Solution steps

Find the slope of

y = 5^{x - 3} : \frac{d y}{d x} = ln (5) \cdot 5^{x - 3}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = ln (5)

Find the line with slope m=

ln (5)

and passing through

(3, 1) : y = ln (5) x + 1 - ln (5) \cdot 3

y = ln (5) x + 1 - ln (5) \cdot 3

f (x) = x^{3} - 2 x^{2} + x + 1

\int_{- 2}^{- 1} \frac{1}{( x - 1 ) ^{3}} d x

7 x + 12, 0, 10

\frac{1}{s ^{3} ( s + 2 )}

\frac{\partial}{\partial x} (x^{2} y + y^{2} z + z^{2} x)

What is the tangent of y=5^{x-3},(3,1) ?
The tangent of y=5^{x-3},(3,1) is y=ln(5)x+1-ln(5)*3