Solve for x and y real: a) 7 - 2yi = 13x + 3i b.) sin x + 1/2i = 1 + (cosy)i solve for x: a.) (log_4_(x+1))^2 - 3log_4_(x+1) - 4 = 0 _4_ is base thanks!!!
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Originally Posted by ^_^Engineer_Adam^_^ Solve for x and y real: a) 7 - 2yi = 13x + 3i rearrange: 13x+2y i = 7 - 3 i, Now as x and y are real equate the real and imaginary terms on each side or the equality to get: 13x=7, and 2y=-3, or x=7/13, and y=-3/2. RonL Please double check the algebra.
Originally Posted by ^_^Engineer_Adam^_^ b.) sin x + 1/2i = 1 + (cosy)i Like the last one: sin(x) - i cos(y)=1 - (1/2) i, so sin(x)=1, and cos(y)=1/2, which you should be able to solve yourself RonL
Originally Posted by ^_^Engineer_Adam^_^ solve for x: a.) (log_4_(x+1))^2 - 3log_4_(x+1) - 4 = 0 _4_ is base put y=log_4 (x+1), then the equation is: y^2-3y-4=0 which has roots y=4, y=-1 So in the first case we have log_4 (x+1)=4, or: x+1=4^4 x=4^4-1=256-1=255. In the second case we have log_4 *x+1)=-1, or x+1=4^-1, x=3/4. RonL Again check the algebra yourself.
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