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Topic: Can you help solve "log [base pi](pi^4cos(pi))"?**Question:**
I am completely stumped where to even begin on this problem.
Again it's log [base pi](pi^4cos(pi))
Here it is if that doesn't make sense
http://www3.wolframalpha.com/input/subpo...
Also, can you explain where you began and why you chose that place? Thanks!

July 19, 2019 / By Andrew

If you just apply the identities you get: log[π](π^4*cos(π)] =log[π]{π^4*(-1)} =-log[π](π^4) =-4log[π](π) =-4 but note log(π^4*cosπ) is not define. Thus this expression is undefine. If you mean log[π]{π^[4cos(π)]} =log[π]{π^(-4)} =-4

👍 224 | 👎 8

Did you like the answer? 4cos(theta) - 2v3 = 0 add 2v3 to both sides to get 4cos(theta) = 2v3 divide both sides by 4 (and reduce) to get cos(theta) = v3/2 determine what values of theta between 0 and 2pi have cosine values of v3/2. theta = 30 degrees or theta = 330 degrees. (or you could use inverse function... but inverse cosine is NOT secant, it is arccos: so theta=arccos(v3/2) will give the reference angle 30degrees, then you must determine the other angle value)

log(base a) a is 1 so log [base pi](pi^4cos(pi)) is 4 cos pi that means 4*(-1)= -4 is the answer

👍 90 | 👎 7

log [base pi](pi^4cos(pi)) = log [base pi] pi^4 + log [base pi] cos(pi) = 4 log [base pi] pi + log [base pi] cos(pi) = 4 + log [base pi] cos(pi) ...

👍 83 | 👎 6

log(base pi)(pi^4cos(pi)) = log(base pi)(pi^4 * (-1)) = log(base pi)(-pi^4) has no real solution If you meant log(base pi)(pi^(4cos(pi))), then log(base pi)(pi^4cos(pi))) = log(base pi)(pi^4(-1)) = log(base pi)(pi^-4) = -4

👍 76 | 👎 5

The answer is not 6 3 log(base 7) 7 = 3 3 log[base 7] 7 + 7^2 log[base 7] 3 - log[base 7] = 3+49*(log(3)/log(7))-(log(3)/log(7)) =30.0996 Here, log = log(base 10)

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