StudyGurl14
  • StudyGurl14
HELP PLEASE Solving for x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
StudyGurl14
  • StudyGurl14
|dw:1440620935256:dw| Obviously, I did something wrong.
StudyGurl14
  • StudyGurl14
@perl
perl
  • perl
|dw:1440621059469:dw|

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More answers

StudyGurl14
  • StudyGurl14
Oh, I see. I think.
perl
  • perl
multiply both sides by 2^x, to get rid of 2^x in the denominator
StudyGurl14
  • StudyGurl14
|dw:1440621096683:dw| Like that?
perl
  • perl
|dw:1440621179162:dw|
freckles
  • freckles
hint: you will wind up with a quadratic in terms of 2^x
perl
  • perl
|dw:1440621199302:dw|
StudyGurl14
  • StudyGurl14
Oh, I forgot the 1.
StudyGurl14
  • StudyGurl14
so...? |dw:1440621240821:dw|
StudyGurl14
  • StudyGurl14
???
StudyGurl14
  • StudyGurl14
Now do I use ln?
perl
  • perl
|dw:1440621466522:dw|
StudyGurl14
  • StudyGurl14
Oh, so 2^2x
perl
  • perl
|dw:1440621493371:dw|
StudyGurl14
  • StudyGurl14
|dw:1440621524603:dw| ???
perl
  • perl
Now you have a quadratic equation.
StudyGurl14
  • StudyGurl14
How?^
phi
  • phi
FYI, 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
  • StudyGurl14
Okay, so how do I distrubite the ln @phi ?
perl
  • perl
if we let u = 2^x. then we have u^2 - 5*u +1 = 0
perl
  • perl
you can't distribute the ln
StudyGurl14
  • StudyGurl14
Oh, I see @perl
StudyGurl14
  • StudyGurl14
so.... 5/2 +- sqrt{21}/2 ?
phi
  • phi
yes, and now you can do 2^x = mess and take the log of both sides
StudyGurl14
  • StudyGurl14
what's "mess" ?
phi
  • phi
5/2 +- sqrt{21}/2
StudyGurl14
  • StudyGurl14
okay, so |dw:1440621917929:dw| like that?
perl
  • perl
we 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
  • StudyGurl14
I got approx 2.26?
perl
  • perl
yes
StudyGurl14
  • StudyGurl14
Okay, thank you.
perl
  • perl
and the other solution?
StudyGurl14
  • StudyGurl14
Do I do it again for the negative one?
perl
  • perl
right
StudyGurl14
  • StudyGurl14
so approx -2.26
perl
  • perl
indeed :)
StudyGurl14
  • StudyGurl14
Thank you!
perl
  • perl
and 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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