Find slope of a tangent to the curve of

[mlj2005]

y=1n(25x+3) at x=2

Answer .5?

 

[tkhunny]

y=1n(25x+3) at x=2

Answer .5?

Why is this a problem? What part of the evaluation causes you consternation? Feel free to give us more to work with.

ln(25*(2)+3) = ln(50+3) = ln(53) = 3.9702919135521218341444691390291

Not very interesting. Are you sure we have the right problem statement?

 

[Gene]

Close enough if you are graphing it. It is actually
25/(25x+3)
You don't say what you are taking, so we don't know what method you are supposed to use.

 

[stapel]

I think the poster isn't actually evaluating y at x = 2, as the post says, but is trying to find the tangent to y at x = 2, if the subject line is to be included as the first half of the sentence.

(This is why, mlj2005, it has been suggested that you include the entire post within the post, rather than hiding part of it in the subject line. As you have seen here and elsewhere, the result might not be what you intended.)

Eliz.

 

[tkhunny]

Ah, I spotted one of two title splits. I missed this one.

If you are doing calculus, please post in the calculus section. It is not possible to tell how to proceed in "Other Math" without very clear direction.