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Help!?!?!?!?
17. Solve 1n 2 + 1n x = 5. Round to the nearest thousandth, if necessary.
 one year ago
 one year ago
Help!?!?!?!? 17. Solve 1n 2 + 1n x = 5. Round to the nearest thousandth, if necessary.
 one year ago
 one year ago

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swiftskier96Best ResponseYou've already chosen the best response.0
I tried to put it in on my calculator, but it doesnt work. Either that, or i dont know how to do it.
 one year ago

swiftskier96Best ResponseYou've already chosen the best response.0
I used an online calculator. I typed the whole thing in like it says above, and it says that its unknown.
 one year ago

artix_17Best ResponseYou've already chosen the best response.1
okay so first you have to group the terms together. Unknown on the left and known values on the right. ln x = 5  ln 2
 one year ago

artix_17Best ResponseYou've already chosen the best response.1
So ln x = \[\log_{e} x\]
 one year ago

artix_17Best ResponseYou've already chosen the best response.1
To find x, you need to move the log base over to the other side Therefore x = e^(5ln 2)
 one year ago

swiftskier96Best ResponseYou've already chosen the best response.0
Oh ok. That makes sence.
 one year ago

artix_17Best ResponseYou've already chosen the best response.1
You can simplify 5ln 2 further using calculator
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
\[\log_{}a + \log b = \log (ab) \]
 one year ago

swiftskier96Best ResponseYou've already chosen the best response.0
I got 4.3068528194400546905827678785418. (But ill obviously have to shorten it. :)
 one year ago

swiftskier96Best ResponseYou've already chosen the best response.0
So it would be x = e^4.3?
 one year ago

swiftskier96Best ResponseYou've already chosen the best response.0
So is that my final answer? Or do i have to solve more of it?
 one year ago

artix_17Best ResponseYou've already chosen the best response.1
though its more accurate to leave it as 5ln 2, that's your final answer
 one year ago

swiftskier96Best ResponseYou've already chosen the best response.0
Oh ok. Cool! Thanks!!
 one year ago
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