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September 3, 2014

September 3, 2014

Posted by **Jo** on Wednesday, January 18, 2012 at 3:00am.

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

- College Algebra -
**Reiny**, Wednesday, January 18, 2012 at 8:41amlog

_{6}[ (64x^3+1)/(4x+1) ] = 1

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

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

16x^2 - 4x - 5 = 0

x= (4 ± √336)/32

= (4 ± 4√21)/32

= (1 ± 21)/8

We have to reject the negative answer or else the log would be undefined

so

x = (1 + √21)/8

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