When is there one solution? - GRE Mathematics Subject Test 

I did this in a slightly different (worse?) way. I solved for x using the Lambert W function and got x = e^(-W(-4 c)/4). The W function isn't defined for arguments < -1/e. It is equal to -1 at -1/e, has two values between -1/e and 0, and has a single value when the argument is greater than zero. Since all of the answer options are positive, the argument to W will be negative - and so it must be -1/e, the only negative argument that yields a single real value, and therefore c must be 1/(4e).

What a nice problem! Really puts into question if the examinee understands the meaning of a derivative!

Derivative if log(x) is not 1/x. Graphing tech confirms that the given solution does not work for this question. If you meant ln(x) = cx^4, then it does work. The way it is wtitten, the correct answer is c = log(e)/(4e)

The answer is correct, but somewhere in the middle of the step is incorrect because the derivative of log(x) is 1/(xln(10)). From there, if you differentiate both sides of the equation, you get 1/(xln(10))=4cx^3, so 4cx^4ln(10)=1. Now substitute cx^4=log(x), which is 4log(x)ln(10)=1. Note that log(x)=ln(x)/ln(10). Basically, 4ln(x)=1, so x=e^(1/4). If you raise both sides by 4, we have x^4=e. Putting back in the equation that was differentiated, we 4ce=1. Solving for c gives us c=1/(4e). Hence, the answer must be choice A.

Please mention the base of the log in the question

Differentiating does not always work

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