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August 31, 2014

August 31, 2014

Posted by **Teddy** on Wednesday, August 11, 2010 at 10:16pm.

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 6:26am2^(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 7:34amWell 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 8:39pmwhats 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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