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Question

Question: Solving \(3-2yi={{9}^{x}}-6i\), for \(x\) and \(y\) real, we get: 1\. \(x=0.5,y=3.5\) 2\. \(x=0....

Solving 32yi=9x6i3-2yi={{9}^{x}}-6i, for xx and yy real, we get:
1. x=0.5,y=3.5x=0.5,y=3.5
2. x=0.5,y=3x=0.5,y=3
3. x=0.5,y=7x=0.5,y=7
4. None of these

Explanation

Solution

To solve the given question in which we have an imaginary equation as 32yi=9x6i3-2yi={{9}^{x}}-6i, we need to compare the real number with real number and coefficient of imaginary number to coefficient of imaginary number. After that, we will do required mathematical calculations to simplify the obtained expression and will find the required values.

Complete step by step answer:
Since, we have the given expression from the question as:
32yi=9x6i\Rightarrow 3-2yi={{9}^{x}}-6i
Now, we will compare the corresponding values and variables of the left hand side to corresponding values and variables of the right hand side one by one. First of all, we will do the comparison for the real part of the expression as:
3=9x\Rightarrow 3={{9}^{x}}
As we know, 33 is the square root of 99. So, we can write it in under root of 99 as:
9=9x\Rightarrow \sqrt{9}={{9}^{x}}
Here, we can write the under root in the form of power as 12\dfrac{1}{2}. So, we can write the above step as:
912=9x\Rightarrow {{9}^{\dfrac{1}{2}}}={{9}^{x}}
Now, the base will be canceled because the base is the same on both sides. So, we will have the above expression as:
x=12\Rightarrow x=\dfrac{1}{2}
Here, we will convert the fraction into decimal as:
x=0.5\Rightarrow x=0.5
Since, we got the value of xx. Now, we will do the comparison of imaginary part of the expression as:
2y=6\Rightarrow 2y=6
Here, we will divide by 22 each side of the above step as:
2y2=62\Rightarrow \dfrac{2y}{2}=\dfrac{6}{2}
After solving it, we will get the value of yy as:
y=3\Rightarrow y=3
Hence, we will have the value of xx and yy as 0.50.5 and 33 respectively.

So, the correct answer is “Option 2”.

Note: A complex number is a number that contains both the real number and imaginary unit within itself. It is expressed as x+iyx+iy, where xx and yy are real numbers and iiis an imaginary unit that satisfies the condition i2=1{{i}^{2}}=-1.