Another Log problem: cos^{-1}(u) + csc^{-1}(v), where u and v are...

[bobrossu]

Problem:

Evaluate cos^-1(u)+csc^-1(v) where u and v are the solution to the system:

ln(v)=-(1/2)ln(u)
2log₂(u)=2log₂(v)=-3

Work so far:

log₂(u²)=log₂(v²)-3 ----> log₂(u²/v²)=-3 ---> 2^-3=(u²/v²) --> (1/8)=(u²/v²)

Not sure what to do next... also confused about how ln rules work compared to log rules.

 

[JeffM]

Problem:

Evaluate cos^-1(u)+csc^-1(v) where u and v are the solution to the system:

ln(v)=-(1/2)ln(u)
2log₂(u)=2log₂(v)=-3

Work so far:

log₂(u²)=log₂(v²)-3 ----> log₂(u²/v²)=-3 ---> 2^-3=(u²/v²) --> (1/8)=(u²/v²)

Not sure what to do next... also confused about how ln rules work compared to log rules.

\(\displaystyle ln(x) = log_e(x)\)

so the ordinary log rules apply to the natural logarithm.

 

[bobrossu]

\(\displaystyle ln(x) = log_e(x)\)

so the ordinary log rules apply to the natural logarithm.

still unsure what to do next, someone help me out.

I'm guessing that I turn the ln(x) into loge(x) form and get loge(v)=log_e(u^(-1/2)) -----> v=u^(-1/2) ------> (u²/u^(-1/2)) -----> u³=(1/8) ---> u=(1/2) and v= root 2

 

[ksdhart2]

still unsure what to do next, someone help me out.

I'm guessing that I turn the ln(x) into loge(x) form and get loge(v)=log_e(u^(-1/2)) -----> v=u^(-1/2) ------> (u²/u^(-1/2)) -----> u³=(1/8) ---> u=(1/2) and v= root 2

Yes, that seems like an excellent way to proceed. As a technical point, there are actually two other answers to the equation u³ = 1/8, although both of them are complex. I don't know what your class's method for dealing with complex numbers thus far has been (many just ignore them altogether...)

For now, we'll work only with the real solution. You've found values for u and v, so now it's time to return to the first part of the question which asks you to evaluate cos^-1(u). So what is the inverse cosine of 1/2? Or, put another way, the cosine of what gives 1/2? Similarly, what is the inverse cosecant of sqrt(2)? The cosecant of what gives sqrt(2)? Note that there will be infinitely many solutions to these questions due to the periodicity of sine and cosine, but you can find a "family" of solutions that gives them all.