# the what is the equation of the curve of the tangent to the curve if x=10?y=log(base 5)x-log(base 5)2+log(base 5)(x-5)

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You need to find the slope of the tangent line, hence you need to differentiate the function with respect to x such that:

`dy/dx = ((ln x)/(ln 5))' - (log_5 2)' + ((ln (x-5))/(ln 5))'`

`f'(x) = dy/dx = 1/(x*ln 5) + 1/((x-5)ln5)`

Plugging x = 10 in the equation of `dy/dx ` yields:

`f'(10) = 1/(10ln5) + 1/(5ln5) = 1/(ln5)(1/10 + 1/5) = 3/(10ln5)`

Hence, the slope of the tangent line is `m =3/(10ln5).`

You need to remember the slope point form of equation of line such that:

`y - f(2) = m(x - 10)`

Plugging x = 10 in the equation of the function yields:

`f(10) = log_5 10 - log_5 2 + log_5 5`

`f(10) = log_5(2*5) - log_5 2 + 1`

Using the properties of logarithms yields:

`f(10) = log_5 2 + log_5 5 - log_5 2 + 1`

Reducing like terms yields:

`f(10) = 2`

Hence, the equation line of tangent is: `y - 2 = 3/(10ln5)(x - 10)`

Opening the brackets yields:

`y = 2 + (3x)/(10ln5) - 3/(ln5)`

`` `y = (3x)/(10ln5) + (ln25 - 3lne)/(ln5)`

`` `y = (3x)/(10ln5) + (ln(25/(e^3)))/(ln5)`

**Hence, the equation of tangent line to curve is `y = (3x)/(10ln5) + (ln(25/(e^3)))/(ln5).` **