Asked by Heather
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
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
Answers
Answered by
Steve
looks good to me.
something bother you about that answer?
something bother you about that answer?
Answered by
Heather
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?
Answered by
Steve
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.
It's just a number.
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