Question
solve :log (12x-10)=1+log2(4x+3)
Answers
charles
I dont understand
Reiny
I interpret your question as
log (12x-10)=1+log (2(4x+3))
then, replace 1 with log 10
log (12x-10)=log 10 +log (8x + 6)))
log (12x-10)=log (10(8x+6))
12x - 10 = 80x + 60
72 = -68x
x = -72/68 = -18/17
but that would make log (12x-10) undefined, so there is no solution
Let me know what you meant by log2(4x+3)
log (12x-10)=1+log (2(4x+3))
then, replace 1 with log 10
log (12x-10)=log 10 +log (8x + 6)))
log (12x-10)=log (10(8x+6))
12x - 10 = 80x + 60
72 = -68x
x = -72/68 = -18/17
but that would make log (12x-10) undefined, so there is no solution
Let me know what you meant by log2(4x+3)
oobleck
maybe you mean to take both logs base 2? If so, then following Reiny's attempt above, that would let us pick up at
log (12x-10)=log 2 +log (8x + 6)))
log (12x-10)=log (2(8x+6))
12x-10 = 16x+12
4x = 22
x = 11/2
Possibly, but I suspect another typo.
Online, using logs to any base, say, 2, many write
log_2(x) with the underscore indicating a subscript (or log base).
Care to retype your equation?
log (12x-10)=log 2 +log (8x + 6)))
log (12x-10)=log (2(8x+6))
12x-10 = 16x+12
4x = 22
x = 11/2
Possibly, but I suspect another typo.
Online, using logs to any base, say, 2, many write
log_2(x) with the underscore indicating a subscript (or log base).
Care to retype your equation?