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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- Oct 2nd 2006, 05:31 AM^_^Engineer_Adam^_^Complex number and Logarithm help me
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!!! - Oct 2nd 2006, 05:55 AMCaptainBlack
- Oct 2nd 2006, 05:58 AMCaptainBlack
- Oct 2nd 2006, 06:08 AMCaptainBlack
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.