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Mathematics 16 Online
OpenStudy (anonymous):

Solve for x - Do not round any intermediate computations, and round your answer to the nearest hundredth. 5-21n x=6 I can send screen shot if it would help.Thanks

OpenStudy (anonymous):

\[5-2*\ln (x)=6\] Is t his your formula ?

OpenStudy (anonymous):

Here is what is on the study material (attached) - Not sure how to handle this problem. Thank you

OpenStudy (anonymous):

forget the ln(x) for a moment and see if you can reduce this form to something that looks like ln(x) = ... Can you do that ?

OpenStudy (anonymous):

Treat this equation as a regular equation and follow the same steps as you would do for an equation like 5 - 2*k = 6. Can you solve this one for k (so rewrite into k = ...)

OpenStudy (anonymous):

k= -1/2 ?

OpenStudy (anonymous):

Correct, well done. Now replace k by ln(x) and you've taken the first step towards solving the original equation. I'm sure you agree that we've now reached: k = ln(x) = -1/2

OpenStudy (anonymous):

The next step is to get rid of the monster ln(x). I'm not sure how comfortable you are with natural logarithms en the natural base 'e'.

OpenStudy (anonymous):

Not comfortable at all to be honest

OpenStudy (anonymous):

breaking this thing down to a "regualr" equation is what i am struggling with

OpenStudy (anonymous):

No surprise, it takes quite some training to not be scared by this. Have you seen anything like e^ln(x) before ?

OpenStudy (anonymous):

no - i am trying to prep for a college algebra class i have coming up in a week or so and haven't made it to anything like that yet

OpenStudy (anonymous):

|dw:1377006944978:dw|OK, sometimes it's unavoidable to 'know' some simple rules. Without going into too many details, the simplest that you may try to remember in the future is in the attachment. These two are pretty helpful and if you take the first one and look at our reduced equation, you may see the answer.

OpenStudy (anonymous):

You already completed the first step, by transfering the rather complex equation to: ln(x) = -1/2 You may also use both sides as powers of e, so transforming the equation to e^ln(x) = e ^-1/2 In this case, you see a resemblance between the left hand side and the formula 1) in my attachment above, true ?

OpenStudy (anonymous):

|dw:1377007241170:dw|

OpenStudy (anonymous):

ok - i plugged it in and that makes sense - sorry took me a minute. THANK YOU for your help. I just found this website as I was struggling going over some practice test.

OpenStudy (anonymous):

No worries, as I said, sometimes it's unavoidable to actually learn these ground rules on top of your head. You will do fine, take care...

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