SOLUTION: Solve foe x: a) (0.5)^(x+2)=3^x b) 1/8(16^(2x-1))=(sqrt2)^x c)2-log(5x)=log(x+1)

Question 138869: Solve foe x:
a) (0.5)^(x+2)=3^x
b) 1/8(16^(2x-1))=(sqrt2)^x
c)2-log(5x)=log(x+1)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a) (0.5)^(x+2)=3^x
(x+2)log(0.5) = xlog3
xlog(0.5) + 2log(0.5) = xlog3
x(log0.5-log3) = -2log(0.5)
x log((1/2)/3) = log(1/2)^(-2)
x log(1/6) = log 4
x = log4 / log(1/6)
x = -0.773706...
----------------------------
b) 1/8(16^(2x-1))=(sqrt2)^x
16^(2x-1) = 8*(sqrt2)^x
2^[4(2x-1)] = 2^3*2^(x/2)
2^(8x-4] = 2^[(x+6)/2]
8x-4 = (x+6)/2
16x-8 = x+6
15x = 14
x = 14/15
---------------------------
c)2-log(5x)=log(x+1)
2 = log(5x)+log(x+1)
2 = log[5x^2+5x]
5x^2+5x=10^2
5x^2+5x-100 = 0
x^2+x-20 = 0
(x+5)(x-4) = 0
Positive solution:
x = 4
===============
Cheers,
Stan H.

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