# Question on logarithmic differentiation and absolute value

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1. Mar 9, 2015

### Mr Davis 97

For the problem of differentiating $y = x^5(3x-1)^3$ using logarithmic differentiation, the solution provides the first step as rewriting the functions as $\left |y \right | = \left | x \right |^5 \cdot \left | 3x-1 \right |^3$. This confuses me. First, how are we, mathematically, able to do this? How can we just select which terms inside of the function to take the absolute value of? Why wouldn't it be $\left |y \right | = \left | x^5(3x-1)^3 \right |$? I would love for someone to explain this to me. Second of all, why is this an essential step in logarithmic differentiation? Why can't we just take the natural log of $y = x^5(3x-1)^3$ and be done with it?

2. Mar 9, 2015

### PWiz

$|a$ x $b|=|a|$ x $|b|$. This is a rule. Additonally, logarithms are only defined for positive values. It is essential to put a modulus because $x$ can take any value and the result can be negative. I also think that an easier way to solve this would be to simply use the product rule for differentiation.