Posted by **Heather** on Friday, January 4, 2013 at 4:11pm.

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

- calculus -
**Steve**, Friday, January 4, 2013 at 4:16pm
looks good to me.

something bother you about that answer?

- calculus -
**Heather**, Friday, January 4, 2013 at 4:49pm
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?

- calculus -
**Steve**, Saturday, January 5, 2013 at 12:18am
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.

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