solve the logarithmic equation. Be sure to reject any value of x that is not in the domain of the orginal logarithmic expression.
In square root x+7=1
I get this far and then get stuck, can you help me?
In square root x+7=1
e=square root x+7
e^2=x+7
x+7=e^2
x=e^2-7
looks good to me.
something bother you about that answer?
should I go farther or is that it. I don't know it that it the final answer--I think that is what is bothering me?
the answer is exact. Going any further would just be to approximate it with a decimal. e^2-7 is just as good as π/4 or √23-ln(3).
It's just a number.
To solve the equation √(x+7) = 1, you're on the right track.
Here are the steps to complete the solution:
1. Start with the equation: √(x+7) = 1.
2. Square both sides of the equation to eliminate the square root: (√(x+7))^2 = 1^2.
3. Simplify the left side by squaring the square root: x + 7 = 1.
4. Now, isolate the variable x by subtracting 7 from both sides of the equation: x = 1 - 7.
5. Simplify the right side: x = -6.
So the solution to the equation √(x+7) = 1 is x = -6.
However, before finalizing the answer, we need to check if this solution is within the domain of the original logarithmic expression.
In the original equation, we have √(x+7), which means the expression inside the square root must be greater than or equal to zero.
So we need to check if x + 7 ≥ 0.
Plugging in x = -6 into x + 7 ≥ 0, we have -6 + 7 ≥ 0, which simplifies to 1 ≥ 0.
Since 1 is indeed greater than or equal to zero, the solution x = -6 satisfies the domain requirement.
Therefore, the final solution to the equation √(x+7) = 1 is x = -6.