Logarithms: using logs to solve 2^x + 2^(-x) = 3

[iheartthemusic29]

2^x+2^-x=3
Can someone explain how to solve this using logarithms?

 

[stapel]

2^x+2^-x=3

Do you have to solve this using logs...? It would be so much simpler to solve by multiplying through by 2[sup:ezfwq2jk]x[/sup:ezfwq2jk], and then using quadratic methods.... :shock:

Eliz.

 

[iheartthemusic29]

Re:

2^x+2^-x=3

Do you have to solve this using logs...? It would be so much simpler to solve by multiplying through by 2[sup:1co3ikgy]x[/sup:1co3ikgy], and then using quadratic methods.... :shock:

Eliz.

I tried that and got to (2^x)^2-3(2^x)+1=0
Then said let u=2^x.
u^2-3u+1=0
However, I then get outrageous decimals that are incorrect, so I obviously messed up somewhere. :?

 

[tkhunny]

Why are you getting ANY decimals? Put down your calculator.

You have it right. \(\displaystyle 2^{x}\;=\;\frac{3 \pm \sqrt{5}}{2}\)

Now the logs. Base 2 seems convenient. \(\displaystyle x\;=\;log_{2}\left(\frac{3 \pm \sqrt{5}}{2}\right)\)

I don't see any obscene decimals.