# P2 logs

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#1
Solve 5^x + 5^-X =6
0
15 years ago
#2
Solve 5^x + 5^-X =6

Call u = 5^x

u + 1/u = 6
u^2 - 5 = 0
u^2 = 6
u = +/- root 6

5^x = +/- root 6
x log 5 = log(+/- root 6)
x = [log(+/- root 6)]/(log 5)

I think this works... not sure.
0
15 years ago
#3
(Original post by droid)
Solve 5^x + 5^-X =6
5^x+1/5^x=6
so (5^x)^2+1=6*5^x
let y=5^x
so y^2-6y+1=0
so (y-3)^2=8
so y=3+-2sqrt2
so then u have to do logs of both sides

5^x=3+2sqrt2
xln5=ln3+2sqrt2)
so x=(ln(3+2sqrt2)/ln5
0
15 years ago
#4
(Original post by mik1a)
Solve 5^x + 5^-X =6

Call u = 5^x

u + 1/u = 6
u^2 - 5 = 0
u^2 = 6
u seem to have multiplied both sides by u but forgotten to do this to the 6 on the r.h.s.
0
15 years ago
#5
(Original post by lgs98jonee)
u seem to have multiplied both sides by u but forgotten to do this to the 6 on the r.h.s.
Oh yeah.

Solve 5^x + 5^-X =6

Call u = 5^x

u + 1/u = 6
u^2 + 1 = 6u
u^2 - 6u + 1 = 0
(u - 3)^2 -8 = 0
(u-3)^2 = 8
u - 3 = +/- 2√2
u = 3 +/- 2√2

etc/ like you said
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