SOLUTION: 1. 4^x+6(4^-x)=5 2. 5(5^2x)+8(5^x)=6

Question 394648: 1. 4^x+6(4^-x)=5
2. 5(5^2x)+8(5^x)=6
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1. 4^x+6(4^-x)=5
2. 5(5^2x)+8(5^x)=6
solving first problem
4^x+6(4^-x)=5
4^x+6(1/4^x)=5
multiply by LCD 4^x
4^2x+6=5(4^x)
4^2x-5(4^x)+6=0
let 4^x = u
then u^2 =4^2x
u^2-5u+6=0
factor to solve
(u-3)(u-2)=0
u=3, u=2
4^x=3
xlog4=log3
x=log3/log4=.79248
4^x=2
xlog4=log2
x=log2/log4=.5000
answers: x=.79248 or .5
check: 4^x+6(4^-x)=5
f(.79248)=4^.79428+6*(1/4^.79248)
=3.00749+2.00000 = 5.00749

Solving second problem
5(5^2x)+8(5^x)=6
let u=5^x
then u^2 =5^2x
5u^2+8u-6=0
Use following quadratic formula to solve
with a=5, b=8,c=-6

u=(-8+-sqrt(8^2-4*5*-6))/10
=(-8+-sqrt(184))/10
=(-8+-13.5646)/10
=-21.5646/10 or 5.5646/10
=-2.15646 or .55646
5^x=-2.15646 (reject,no solution,(5^x)>0
5^x=.55646
xlog5=log.55646
x=log.55646/log5
ans: x=-.36420
check:
5(5^2x)+8(5^x)=6
1.54824+4.45169 =5.99993

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