Solve the System of Equations 2^(x+y)=16 2^(2x+y)=1/8
This problem is asking for a solution to a system of two equations where the variables x and y are exponents of the base 2. The system comprises two different equations which must both be true for the solution. The first equation indicates that the sum of the exponents x and y, when used as powers of 2, equals 16. The second equation states that the combination of the exponents, with twice the value of x added to y, when used as powers of 2, equals 1/8. The task is to determine the values of the variables x and y that satisfy both equations simultaneously.
$2^{x + y} = 16$$2^{2 x + y} = \frac{1}{8}$
Answer
Solution:
Step 1: Isolate $x$ in the first equation $2^{x + y} = 16$.
Express both sides of the equation with the same base: $2^{x + y} = 2^4$.
Equate the exponents since the bases are equal: $x + y = 4$.
Rearrange to solve for $x$: $x = 4 - y$.
Step 2: Substitute $x$ with $4 - y$ in the second equation $2^{2x + y} = \frac{1}{8}$.
Insert $4 - y$ in place of $x$: $2^{2(4 - y) + y} = \frac{1}{8}$.
Expand and simplify the exponent:
Apply the distributive property: $2^{8 - 2y + y} = \frac{1}{8}$.
Combine like terms: $2^{8 - y} = \frac{1}{8}$.
Step 3: Solve for $y$ in the equation $2^{8 - y} = \frac{1}{8}$.
Use the negative exponent rule to rewrite $\frac{1}{8}$: $2^{8 - y} = 2^{-3}$.
Since the bases are equal, set the exponents equal to each other: $8 - y = -3$.
Solve for $y$:
Isolate $y$: $-y = -3 - 8$.
Simplify: $-y = -11$.
Divide by $-1$ to get $y$: $y = 11$.
Step 4: Find $x$ by substituting $y = 11$ back into $x = 4 - y$.
Replace $y$ with $11$: $x = 4 - 11$.
Simplify to find $x$: $x = -7$.
Step 5: Write the solution as an ordered pair.
- The solution is the pair of values for $x$ and $y$: $(-7, 11)$.
Step 6: Present the result in different forms.
Point Form: $(-7, 11)$.
Equation Form: $x = -7, y = 11$.
Knowledge Notes:
To solve a system of equations where the equations are in the form of exponents with the same base, we can use the property that if $a^m = a^n$, then $m = n$ provided that $a$ is not equal to $0$ or $1$. This allows us to equate the exponents and solve for the variables algebraically.
In this problem, we used the following steps:
Isolation of a Variable: We first isolated one variable, $x$, in one of the equations.
Substitution: We substituted the expression for $x$ into the other equation to solve for $y$.
Simplification: We simplified the equations by applying algebraic rules such as the distributive property and combining like terms.
Negative Exponent Rule: We used the rule that $a^{-n} = \frac{1}{a^n}$ to rewrite expressions with negative exponents.
Solving for the Second Variable: We solved for the second variable, $y$, by equating the exponents with the same base.
Back-Substitution: We substituted the value of $y$ back into the expression for $x$ to find its value.
Solution Representation: We represented the solution as an ordered pair and in equation form.
These steps are a systematic approach to solving systems of exponential equations and can be applied to similar problems.
