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# What are all numerical solutions to the equation 2cos^2(ln(3x))+cos(ln(3x))-1=0

Hello, Diana --

This will involve a u-substitution and a factored quadratic.

Let u = cos(ln(3x))

Then 2u+ u - 1 = 0

Factor:     (2u - 1)(u + 1) = 0

2u - 1  =  0    or   u + 1  = 0

u   =  1/2           u  =  -1

Resubstitute:

cos(ln(3x))  = 1/2      cos(ln(3x)) = -1

ln(3x)  =  pi/3                 ln(3x)  = pi

ln 3 + ln x   = pi/3          ln 3 + ln x  = pi

ln x = pi/3 - ln 3              ln x =  pi - ln 3

x  =  e(pi/3 - ln 3)              x  = e(pi - ln 3)

There may be other solutions that are integer multiples of the pi/3 and pi compnents.

I'd prefer to have the answer as e^(pi/3)/3 and e^(pi)/3, but otherwise good answer. Diana - you can get that from either using exponent rules on the last step, (ab-c = a/ ac) or using eln(3x) = 3x a few steps earlier.

If you don't split ln(3x) = ln3 + lnx, you can get answer quicker.

ln(3x) = pi/3

3x = epi/3

x = (1/3)epi/3

let y = cos(ln(3x)) then by substutution we get, 2y^2 +y -1=0, so by factoring out we get

1/2 *(2y+2)*(2y-1) = 0 or (y+1)*(2y-1)=0, so y= -1 or y= 1/2. now let v = ln(3x), so we get

v = arccosine(-1) = pi or v= arccosine(1/2)=pi/3. so x = (1/3)e^(pi) or x= (1/3) e^(pi/3)

Since there is no restriction on the domain, you probably need to find all the solutions.

2cos2(ln(3x))+cos(ln(3x))-1=0