What is the tangent of y=log_{2}(x),(32,5) ?

The tangent of y=log_{2}(x),(32,5) is y= 1/(32ln(2))x+5-1/(ln(2))

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tangent of y=log_{2}(x),(32,5)

tangent of

y = lo g_{2} (x), (32, 5)

y = \frac{1}{32 ln ( 2 )} x + 5 - \frac{1}{ln ( 2 )}

Solution steps

Find the slope of

y = lo g_{2} (x) : \frac{d y}{d x} = \frac{1}{x ln ( 2 )}

Compute the slope of

y = lo g_{2} (x)

(32, 5) : m = \frac{1}{32 ln ( 2 )}

Find the line with slope m=

\frac{1}{32 ln ( 2 )}

and passing through

(32, 5) : y = \frac{1}{32 ln ( 2 )} x + 5 - \frac{1}{ln ( 2 )}

y = \frac{1}{32 ln ( 2 )} x + 5 - \frac{1}{ln ( 2 )}

\int (x^{3} + x) d x

\int (18 x^{2} + 3) (6 x^{3} + 3 x)^{6} d x

\int \frac{2 x + 1}{( x + 1 ) ^{3} \cdot ( x ^{2} + 4 ) ^{2}} d x

f (x) = 4^{x}

y = x^{4 x}

What is the tangent of y=log_{2}(x),(32,5) ?
The tangent of y=log_{2}(x),(32,5) is y= 1/(32ln(2))x+5-1/(ln(2))