# (27^cosx)^sinx=3^3cosX/2

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### 3 Answers

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.` **

and this

2x+1 - (21x+39)/(x^2+x-2)>=1/(x+2)

Presentation of the problem is not very clear. I assume that the given problem is:

`(27^cosx)^sinx=3^((3cosx)/2)`

`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...

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Yes! I forgot to mention,

cosx=0, should lead to another set of solutions for x:

`rArr` `x = (2k+1)pi/2`