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September 30, 2014

September 30, 2014

Posted by **Jennifer** on Tuesday, January 3, 2012 at 10:11pm.

log[6](64x^3+1)-log[6](4x+1)=1

- logarithms -
**Reiny**, Tuesday, January 3, 2012 at 10:47pmlog[6] ( (64x^3 + 1)/(4x+1) ) = 1

(64x^3+1)/(4x+1) = 6^1

I can reduce the left side ...

(4x+1)(16x^2 -4x+1)/(4x+1) = 6

16x^2 - 4x - 5 = 0

x = (4 ± √336)/32

but in log (4x+1) , 4x+1 > 0

x > -1/4, so we have to reject the negative x value

x = (4 + √336)/32

= (4 + 4√21)/32

= (1 + √21)/8

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