MHB How would you simplify this logarithmic expression?

[paulmdrdo1]

how would you simplify this

$\displaystyle \frac{1}{5}\ln|\sin5x|+\frac{1}{5}\ln|\csc5x-\cot5x|$

please explain.

 

[Ackbach]

Re: simplification of expression

What have you tried?

 

[paulmdrdo1]

Re: simplification of expression

What have you tried?

i just factored out the 1/5 and I'm stucked.

 

[Fantini]

Re: simplification of expression

Hello Paul! Notice you can factor the $1/5$, yielding $$\frac{1}{5} \left[ \ln |\sin 5x | + \ln |\csc 5x - \cot 5x| \right],$$ and from here use that $\ln a + \ln b = \ln (ab)$. Consequently

$$\frac{1}{5} \left[ \ln |\sin 5x | + \ln |\csc 5x - \cot 5x| \right] = \frac{1}{5} \left[ \ln |\sin (5x) \cdot (\csc (5x) - \cot (5x) )| \right] = \frac{1}{5} \left[ \ln |1 - \cos (5x)| \right].$$

Cheers. :D

 

[soroban]

Re: simplification of expression

Hello, paulmdrdo!

\text{Simplify: }\:\tfrac{1}{5}\ln|\sin5x|+\tfrac{1}{5}\ln|\csc5x-\cot5x|

\begin{array}{ccc}\text{Factor:} & \tfrac{1}{5}\big(\ln|\sin5x| + \ln|\csc5x - \cot5x|\big) \\ \\ \text{Combine logs:} & \tfrac{1}{5}\ln|\sin5x(\csc5x - \cot5x)| \\ \\ \text{Distribute:} & \tfrac{1}{5}\ln|\sin5x\csc5x - \sin5x\cot5x| \\ \\ \text{Simplify:} & \tfrac{1}{5}\ln|1-\cos5x| \\ \\ \text{Further?} & \tfrac{1}{5}\ln\left|2\sin^2\! \tfrac{5x}{2}\right| \end{array}

 

[kaliprasad]

Re: simplification of expression

Hello, paulmdrdo!


\begin{array}{ccc}\text{Factor:} & \tfrac{1}{5}\big(\ln|\sin5x| + \ln|\csc5x - \cot5x|\big) \\ \\ \text{Combine logs:} & \tfrac{1}{5}\ln|\sin5x(\csc5x - \cot5x)| \\ \\ \text{Distribute:} & \tfrac{1}{5}\ln|\sin5x\csc5x - \sin5x\cot5x| \\ \\ \text{Simplify:} & \tfrac{1}{5}\ln|1-\cos5x| \\ \\ \text{Further?} & \tfrac{1}{5}\ln\left|2\sin^2\! \tfrac{5x}{2}\right| \end{array}

In the last step we can remove modulo sign as it positive