Question - Solving a System of Exponential Equations
Solution:
The image shows a system of equations that needs to be solved:$$2^{x+y} = 8$$ and $$2^{3x-y} = 128$$For the first equation, we can rewrite 8 as $$2^3$$ since $$2^3 = 8$$. This gives:$$2^{x+y} = 2^3$$From the property of exponents that states if $$a^m = a^n$$ then $$m=n$$ for any non-zero base $$a$$, we get:$$x + y = 3$$ (Equation 1)For the second equation, we can rewrite 128 as $$2^7$$, since $$2^7 = 128$$. This gives:$$2^{3x-y} = 2^7$$Using the same property of exponents, we get:$$3x - y = 7$$ (Equation 2)Now we have a system of two linear equations in two variables ($$x$$ and $$y$$):1. $$x + y = 3$$2. $$3x - y = 7$$We can solve this system by adding Equation 1 and Equation 2 together to eliminate $$y$$:$$x + y + 3x - y = 3 + 7$$$$4x = 10$$$$x = 10/4$$$$x = 2.5$$With $$x$$ found, we can substitute it back into Equation 1 to find $$y$$:$$2.5 + y = 3$$$$y = 3 - 2.5$$$$y = 0.5$$Therefore, the solution to the system is $$x = 2.5$$ and $$y = 0.5$$.
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