Math Problem Statement
The given system of equations is: log_x 8 = y + 1 and x^y = 6 - x.
Solution
The given system of equations is:
\log_x 8 = y + 1 \\ x^y = 6 - x \end{cases}$$ Let's solve this system step-by-step. ### Step 1: Solve for $$y$$ from the first equation The first equation is: $$\log_x 8 = y + 1$$ Rewrite this in exponential form: $$x^{y+1} = 8$$ Since $$8 = 2^3$$, we have: $$x^{y+1} = 2^3$$ ### Step 2: Solve for $$x$$ Using the second equation: $$x^y = 6 - x$$ We have a system of two equations now: 1. $$x^{y+1} = 8$$ 2. $$x^y = 6 - x$$ Let's substitute $$x^{y+1}$$ from the first equation and solve the system together. We can isolate one variable and try to find values for both $$x$$ and $$y$$. Let me calculate this.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Logarithms
Exponential Equations
System of Equations
Formulas
log_b(a) = c ↔ b^c = a
Exponential form: x^{y+1} = 8
x^y = 6 - x
Theorems
Change of Base Theorem
Laws of Exponents
Suitable Grade Level
Grades 10-12