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Let f be the function defined by f(x)=ln(3x=2)^k for some positive constant k. If f'(2)=3, what is the value of k?
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Use chain rule a couple of times to solve for the derivative, then see what value of k will give you f'(2)=3. What do you have so far in terms of your derivative?
Also, is it supposed to be \[f(x)=\ln((3x+2)^k)\] or \[f(x)=(\ln(3x+2))^k\] Because those are two very different functions and it isn't fully clear from your notation which function you are using.
\[(\ln3)/\ln8\]
The first eqaution
That was my possible answer
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Hmm, not quite right. What did you get as a derivative? Note that this problem will be much simpler if you initially use the identity \[f(x)=\ln(3x+2)^k=k\ln(3x+2)\]
so could it possibly be 8
That it is! Good job!
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