Question
what the solution to the equation below
2^x / x-2 - x+3 / x-4 = x-4 / x^2 - 6x +8
2^x / x-2 - x+3 / x-4 = x-4 / x^2 - 6x +8
Answers
GPT-4o mini
To solve the equation
\[
\frac{2^x}{x - 2} - \frac{x + 3}{x - 4} = \frac{x - 4}{x^2 - 6x + 8}
\]
let's first simplify the right-hand side. Notice that the denominator \(x^2 - 6x + 8\) can be factored:
\[
x^2 - 6x + 8 = (x - 2)(x - 4)
\]
So we can rewrite the right-hand side:
\[
\frac{x - 4}{(x - 2)(x - 4)}
\]
This simplifies to:
\[
\frac{1}{x - 2}
\]
Thus, the equation becomes:
\[
\frac{2^x}{x - 2} - \frac{x + 3}{x - 4} = \frac{1}{x - 2}
\]
We can combine the terms on the left-hand side by getting a common denominator, which is \((x - 2)(x - 4)\):
Starting with:
\[
\frac{2^x (x - 4) - (x + 3)(x - 2)}{(x - 2)(x - 4)} = \frac{1}{x - 2}
\]
Multiplying both sides by \((x - 2)(x - 4)\):
\[
2^x (x - 4) - (x + 3)(x - 2) = (x - 4)
\]
Expanding and simplifying the left side:
\[
2^x (x - 4) - (x^2 - 2x + 3x - 6) = x - 4
\]
\[
2^x (x - 4) - (x^2 + x - 6) = x - 4
\]
\[
2^x (x - 4) - x^2 - x + 6 = x - 4
\]
Bringing everything to one side:
\[
2^x (x - 4) - x^2 - 2x + 10 = 0
\]
To find the solution to this equation, we can substitute possible integer values for \(x\).
Through testing values, we find:
1. For \(x = 2\):
\[
2^2(2 - 4) - 2^2 - 2(2) + 10 = 4(-2) - 4 - 4 + 10 = -8 - 4 - 4 + 10 = -6 \text{ (not a solution)}
\]
2. For \(x = 3\):
\[
2^3(3 - 4) - 3^2 - 2(3) + 10 = 8(-1) - 9 - 6 + 10 = -8 - 9 - 6 + 10 = -13 \text{ (not a solution)}
\]
3. For \(x = 4\):
\[
2^4(4 - 4) - 4^2 - 2(4) + 10 = 16(0) - 16 - 8 + 10 = 0 - 16 - 8 + 10 = -14 \text{ (not a solution)}
\]
Continuing this check until we identify possible solutions or assess behavior around integer points, we ultimately need numerical or graphical methods to finalize solutions, particularly if solutions are not straightforward.
However, it appears through inspection that solutions like \(x = 0\) may lead to integers.
Please check further for assistance or computational help to solve:
\[
2^x (x - 4) = x^2 + 2x - 10
\]
for actual roots numerically or via calculus methods for in-depth analysis. Solutions may exist outside integers or not be solvable in elementary forms.
\[
\frac{2^x}{x - 2} - \frac{x + 3}{x - 4} = \frac{x - 4}{x^2 - 6x + 8}
\]
let's first simplify the right-hand side. Notice that the denominator \(x^2 - 6x + 8\) can be factored:
\[
x^2 - 6x + 8 = (x - 2)(x - 4)
\]
So we can rewrite the right-hand side:
\[
\frac{x - 4}{(x - 2)(x - 4)}
\]
This simplifies to:
\[
\frac{1}{x - 2}
\]
Thus, the equation becomes:
\[
\frac{2^x}{x - 2} - \frac{x + 3}{x - 4} = \frac{1}{x - 2}
\]
We can combine the terms on the left-hand side by getting a common denominator, which is \((x - 2)(x - 4)\):
Starting with:
\[
\frac{2^x (x - 4) - (x + 3)(x - 2)}{(x - 2)(x - 4)} = \frac{1}{x - 2}
\]
Multiplying both sides by \((x - 2)(x - 4)\):
\[
2^x (x - 4) - (x + 3)(x - 2) = (x - 4)
\]
Expanding and simplifying the left side:
\[
2^x (x - 4) - (x^2 - 2x + 3x - 6) = x - 4
\]
\[
2^x (x - 4) - (x^2 + x - 6) = x - 4
\]
\[
2^x (x - 4) - x^2 - x + 6 = x - 4
\]
Bringing everything to one side:
\[
2^x (x - 4) - x^2 - 2x + 10 = 0
\]
To find the solution to this equation, we can substitute possible integer values for \(x\).
Through testing values, we find:
1. For \(x = 2\):
\[
2^2(2 - 4) - 2^2 - 2(2) + 10 = 4(-2) - 4 - 4 + 10 = -8 - 4 - 4 + 10 = -6 \text{ (not a solution)}
\]
2. For \(x = 3\):
\[
2^3(3 - 4) - 3^2 - 2(3) + 10 = 8(-1) - 9 - 6 + 10 = -8 - 9 - 6 + 10 = -13 \text{ (not a solution)}
\]
3. For \(x = 4\):
\[
2^4(4 - 4) - 4^2 - 2(4) + 10 = 16(0) - 16 - 8 + 10 = 0 - 16 - 8 + 10 = -14 \text{ (not a solution)}
\]
Continuing this check until we identify possible solutions or assess behavior around integer points, we ultimately need numerical or graphical methods to finalize solutions, particularly if solutions are not straightforward.
However, it appears through inspection that solutions like \(x = 0\) may lead to integers.
Please check further for assistance or computational help to solve:
\[
2^x (x - 4) = x^2 + 2x - 10
\]
for actual roots numerically or via calculus methods for in-depth analysis. Solutions may exist outside integers or not be solvable in elementary forms.