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Mathematics 14 Online
OpenStudy (linda):

solve for x to the nearest thousandth: 100 ln (2x-3)-10=102

OpenStudy (anonymous):

so is the equation 100(2x-3)-10=102?

OpenStudy (linda):

yes

OpenStudy (anonymous):

So no ln?

OpenStudy (linda):

wait theres an ln in it before the 2x-3

OpenStudy (linda):

sorry, the ln is there

OpenStudy (anonymous):

Use math properties to isolate x. Start by adding 10 to both sides, then divide by 100 on both sides etc.

OpenStudy (linda):

okay, then what should i do with the ln?

OpenStudy (linda):

the ln part confuses me, im not too good with it

OpenStudy (anonymous):

ok, so first im pretty sure that you have to distribute the 100 to the 2 and -3 then you should put the -300 and the -10 together

OpenStudy (anonymous):

making -310, with this you should add it to the other side of the = sign

OpenStudy (anonymous):

e is the inverse of ln. So if both sides are e raised to a power. You will get e^(ln(2x-3)=e^1.12

OpenStudy (linda):

but how do i solve for the x?

OpenStudy (anonymous):

100(2x-3)-10=102 200x-300-10=102 200x-310=102 200x=412 because you added the 210 the the 102 then you should divide 400 by 200 that should leave you with x=2.06

OpenStudy (linda):

you left out the ln (natural logarithm)

OpenStudy (anonymous):

So e^(ln) is 1. So the left side goes to 2x-3. So you should get 2x-3=e^(1.12)

OpenStudy (linda):

ok, so i set 2x-3=1.12. then i added 3 to 1.12 and got 2x=4.12.

OpenStudy (linda):

is that right?

OpenStudy (anonymous):

No. You left out the e. E is a function in itself just like ln. So you can use a calculator to find the value of e^1.12. Then you can continue with your process.

OpenStudy (linda):

ok, so i got 3.065 for e^1.12. then i added 3 to that. i got 2x=6.065. then i divided by two and got x=3.033. sound right?

OpenStudy (anonymous):

Yes!

OpenStudy (anonymous):

You can always check by plugging your value of x back into the equation and seeing if it's equal.

OpenStudy (linda):

yay!! thank you so much!! :)

OpenStudy (anonymous):

No problem. :)

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