solve the logarithmic equation. Be sure to reject any value of x that is not in the domain.
In square root x+7=1
x= ?
What is the decimal approximation to the solution?
x= ?
so sqrt { x+7} = 1 right?
if so... then square both sides to get x+7 =1
x = -6
I think Knights missed the "ln" in front
ln √(x+7) = 1
√(x+7) = e^1 = e
square both sides
x+7 = e^2
x = e^2 - 7
x = appr .389
To solve the logarithmic equation √(x + 7) = 1, we need to isolate the variable x.
1. Start by subtracting 1 from both sides of the equation: √(x + 7) - 1 = 0.
2. Next, square both sides of the equation to eliminate the square root: (√(x + 7) - 1)^2 = 0^2.
3. Expanding the squared term gives us (x + 7) - 2√(x + 7) + 1 = 0.
4. Simplifying further, we get x + 8 - 2√(x + 7) = 0.
5. Move the terms involving the square root to the left side of the equation: -2√(x + 7) = -x - 8.
6. Divide both sides of the equation by -2 to isolate the square root: √(x + 7) = (x + 8) / 2.
7. Now, square both sides of the equation again to eliminate the square root: (√(x + 7))^2 = ((x + 8) / 2)^2.
8. Simplifying yields x + 7 = (x + 8)^2 / 4.
9. Multiply both sides of the equation by 4 to remove the fraction: 4(x + 7) = (x + 8)^2.
10. Expanding and simplifying the equation gives us 4x + 28 = x^2 + 16x + 64.
11. Rearrange the equation to bring all terms to one side: x^2 + 16x + 64 - 4x - 28 = 0.
12. Combine like terms: x^2 + 12x + 36 = 0.
13. Factor the quadratic equation: (x + 6)(x + 6) = 0.
14. Set each factor equal to zero to find possible solutions: x + 6 = 0.
15. Solve for x: x = -6.
Finally, we found that x = -6 satisfies the logarithmic equation.
To find the decimal approximation of x, we simply have x = -6 as the solution.