Posted by Teddy on Wednesday, August 11, 2010 at 10:16pm.
log(base 2)csc(x) = log(base 4)csc(4x^2)
I tried doing the log(base a)b = logb/loga but I can't get past that because I'm not sure what to do after that. How do I set it up so I get like cscx=? or something so I'm able to solve for multiple values of x?
the answer should be pi/3 and 5pi/3

algebra 2  drwls, Thursday, August 12, 2010 at 6:26am
2^(cscx) = (2^2)^[csc(4x^2)]
= 2^[2csc(4x^2)]
This requires that
cscx = 2csc(4x^2)
I don't see an easy way to solve for x.
If x = pi/3
cscx = 1.1547
2 csc(4x^2) = 2.1112
I don't agree with the pi/3 answer.
Make sure the problem is correctly stated.

algebra 2  Teddy, Thursday, August 12, 2010 at 7:34am
Well the original problem was the following but I worked it out to what I put here before:
log(base 2)cotx  2log(base 4)csc2x = log(base 2)cosx
I switched the cosx and the csc2x and used properties of logs to get log(base 2)cscx, but maybe I'm wrong in doing that. is there a better way to work this out from the original above?

algebra 2  eddie, Thursday, March 17, 2011 at 8:39pm
whats the answer fro log base 5 x to the 10th power minus log base 5 x tothe 6th power is equal to 21?
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