Determine the equation of the tangent to the graph of y=log(lnk). My answer: 1/(lnkln10) just can someone confirm this plz
Last edited by john-1; May 2nd 2010 at 07:03 AM.
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Your solution for the equation of the tangent line is correct. The $\displaystyle x$-intercept is the value of $\displaystyle x$ when $\displaystyle y=0$. In our case, $\displaystyle 0=kx_0+\ln\frac{1}{k}-1.$ Hope this helps!
so then how do i solve for k?
because the x is there
Originally Posted by john-1 so then how do i solve for k? hi kx - ln k -1=0 $\displaystyle x=\frac{1+\ln k}{k}>0$ so k would be k<0 or k>1/e but k<0 is not acceptable because ln (<0) is undefined . Do the same to determine the set of values of k so that x is negative .
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