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Mathematics 20 Online
OpenStudy (anonymous):

ln√(x+4) - 2 = -1, solve for x

OpenStudy (anonymous):

Do you think you know what the answer might be or how to get it?

OpenStudy (anonymous):

I think I have an idea and have an answer, but I don't know if it is correct.

OpenStudy (raden):

first, add by 2 to both sides, will giving us ln√(x+4) - 2 = -1 ln√(x+4) - 2 + 2 = -1 + 2 ln√(x+4) = 1

OpenStudy (anonymous):

Have that.

OpenStudy (anonymous):

And what is your idea?

OpenStudy (anonymous):

Then I divided both sides by "e" canceling out the ln and getting me .36788 on the other side.

OpenStudy (anonymous):

Wait why are you dividing by e?

OpenStudy (anonymous):

This is new to me. I thought that the ln is equal to e - .271828 Am I wrong?

OpenStudy (raden):

then setting 1 to ln e now, u have an equation ln√(x+4) = 1 ln√(x+4) = ln e cancel out "ln" √(x+4) = e solve for x

OpenStudy (anonymous):

Oh, thank you. That makes more sense. Got it.

OpenStudy (raden):

you're welcome

OpenStudy (anonymous):

3.3891 should be my answer for x. Correct?

OpenStudy (raden):

from √(x+4) = e squares both sides : x+4 = e^2 subtract by -4 x+4-4 = e^2 - 4 x = e^2 - 4

OpenStudy (raden):

i mean subtract by 4 *

OpenStudy (mertsj):

Time out.

OpenStudy (mertsj):

The base is e. The exponent is 1 so we have: \[e^1=\sqrt{x+4}\]

OpenStudy (mertsj):

Square both sides: \[e^2=x+4\]

OpenStudy (anonymous):

So, can I substitute 2.71828 for e and give a numerical answer for x? If you have time for another one, I'd love help. :) \[e^\ln x^2 - 5^\log(5) 1\]

OpenStudy (mertsj):

Subtract 4 from both sides.

OpenStudy (anonymous):

That gives me 3.3891 right?

OpenStudy (mertsj):

\[x=e^2-4=3.38\]

OpenStudy (mertsj):

Yes

OpenStudy (anonymous):

:)

OpenStudy (raden):

u mean like this, for #2 |dw:1365474768137:dw|

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