Asked by Jennifer
Solve the logarithm equation. Express the solution in exact form only.
log[6](64x^3+1)-log[6](4x+1)=1
log[6](64x^3+1)-log[6](4x+1)=1
Answers
Answered by
Reiny
log[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
(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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