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StudyGurl14
 one year ago
HELP PLEASE
Solving for x.
StudyGurl14
 one year ago
HELP PLEASE Solving for x.

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StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440620935256:dw Obviously, I did something wrong.

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I see. I think.

perl
 one year ago
Best ResponseYou've already chosen the best response.2multiply both sides by 2^x, to get rid of 2^x in the denominator

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440621096683:dw Like that?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1hint: you will wind up with a quadratic in terms of 2^x

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I forgot the 1.

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0so...? dw:1440621240821:dw

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440621524603:dw ???

perl
 one year ago
Best ResponseYou've already chosen the best response.2Now you have a quadratic equation.

phi
 one year ago
Best ResponseYou've already chosen the best response.0FYI, what you did wrong is that when you start with \[ 2^x + 2^{x}= 5 \] we take the ln of both sides \[ \ln \left( 2^x + 2^{x}\right) = \ln 5\] and that is *not* \[ \ln \left( 2^x\right) + \ln \left( 2^{x}\right) = \ln 5\]

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so how do I distrubite the ln @phi ?

perl
 one year ago
Best ResponseYou've already chosen the best response.2if we let u = 2^x. then we have u^2  5*u +1 = 0

perl
 one year ago
Best ResponseYou've already chosen the best response.2you can't distribute the ln

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0so.... 5/2 + sqrt{21}/2 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.0yes, and now you can do 2^x = mess and take the log of both sides

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0okay, so dw:1440621917929:dw like that?

perl
 one year ago
Best ResponseYou've already chosen the best response.2we want to solve (2^x)^2  5*(2^x) +1 = 0 if we let u = 2^x. then we have u^2  5*u +1 = 0 the solutions are u = 5/2 + sqrt{21}/2 , 5/2  sqrt{21}/2 Now back substitute since we really wanted to solve for x 2^x = 5/2 + sqrt{21}/2 2^x = 5/2  sqrt{21}/2 now take ln of both sides ln (2^x)= ln( 5/2 + sqrt{21}/2)

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0I got approx 2.26?

StudyGurl14
 one year ago
Best ResponseYou've already chosen the best response.0Do I do it again for the negative one?

perl
 one year ago
Best ResponseYou've already chosen the best response.2and as phi pointed out, in general ln( x + y) does not equal ln(x) + ln (y) same with other many other functions sin ( x + y) does not equal sin(x) + sin(y),
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