Solving 3 - 2yi = 9^x - 7i,where i^2 = -1,for | vedclass

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(A) Given the equation $3 - 2yi = 9^x - 7i$. Equating the real and imaginary parts on both sides: For the real part: $9^x = 3$. Since $9^x = (3^2)^x =

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$x = 0.5, y = 3.5$

Vedclass · Answer

(A) Given the equation $3 - 2yi = 9^x - 7i$. Equating the real and imaginary parts on both sides: For the real part: $9^x = 3$. Since $9^x = (3^2)^x = 3^{2x}$,we have $3^{2x} = 3^1$. Comparing the exponents,$2x = 1$,which gives $x = 0.5$. For the imaginary part: $-2y = -7$. Dividing both sides by $-2$,we get $y = \frac{7}{2} = 3.5$. Thus,the solution is $x = 0.5$ and $y = 3.5$.

Solving $3 - 2yi = 9^x - 7i$,where $i^2 = -1$,for real values of $x$ and $y$,we get:

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