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You should use the following exponential law, such that:
`(a^x)^y = a^(x*y)`
Reasoning by analogy yields:
`(27^(cos x))^sin x = 27^(cos x*sin x)`
Since `27 = 3^3` yields:
`27^(cos x*sin x) = 3^3^(cos x*sin x) = 3^(3cos x*sin x)`
Equating both sides yields:
`3^(3cos x*sin x) = 3^((3cos x)/2)`
Equating the powers yields:
`3cos x*sin x = (3cos x)/2=> 3cos x*sin x - (3cos x)/2 = 0`
Factoring out `3 cos x` yields:
`3cos x*(sin x - 1/2) = 0`
Using zero product rules yields:
`cos x = 0= > x = +-pi/2 + 2n*pi`
`sin x - 1/2 = 0 => sin x = 1/2 => x = (-1)^n*(pi/6) + n*pi`
Hence, evaluating the solutions to the given equation, using exponential laws, yields `x = +-pi/2 + 2n*pi` and `x = (-1)^n*(pi/6) + n*pi.`
Posted by sciencesolve on June 2, 2013 at 2:43 PM (Answer #2)
Presentation of the problem is not very clear. I assume that the given problem is:
`rArr` `3^3^cosx^sinx = 3^((3cosx)/2)`
`rArr` `3^(3cosxsinx) = 3^((3cosx)/2)`
``Equating the exponents of 3 on both sides,
`3cosxsinx = (3cosx)/2`
dividing both sides by 3cosx,
`rArr` `sinx = 1/2 = sin(pi/6)`
`rArr` `x = pi/6` and `(pi-pi/6)` , i.e. `(5pi)/6` (because sin is positive in 1st as well as 2nd quadrant).
As sin is a periodic function with period `2pi` , the general solutions for x will be:
`x = (pi/6+2kpi)` and `((5pi)/6 + 2kpi)` where, k=0, 1, 2...
Posted by llltkl on June 2, 2013 at 2:34 PM (Answer #1)
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