Logarithm question?

Question

How can I write this out as a single logarithm?

log (x^2 - 9) - log (x^2 + 7x + 12)

Thanks in advance...

Answer 1

log (x^2 - 9) - log (x^2 + 7x + 12)
log [ (x^2 - 9) / (x^2 + 7x + 12) ]
log [ ( (x+3) (x-3) ) / ( (x+3) (x+4) ) ]
cancel the x+3 on top and bottom
log [ (x-3) / (x+4) ] where -4
it can also be written as log (x-3) - log (x+4) to show off to your teacher.

Answer 2

formula is log(A/B) = log A - log B

or log A - log B = log(A/B)

so, by applying theorem,

u get log(x^2-9) - log (x^2 + 7x + 12)

= log ( (x^2-9)/ ( x^2 + 7x+12) )

x^2-9 = (x+3) ( x-3)

x^2+ 7x + 12 = (x+4) (x+3)

so, now u get log ( (x^2-9) / ( x^2+7x+12) )

= log ( ( x-3) / (x+4) )
= log (x-3) - log (x+4)

Answer 3

log (x^2 - 9) - log (x^2 + 7x + 12)
= log [(x^2 - 9) / (x^2 + 7x + 12)]
= log [(x-3)(x+3) / (x+4)(x+3)]
= log [(x-3)/(x+4)]

Note: a^2 - b^2 = (a-b)(a+b), hence x^2-9 = (x-3)(x+3)

Answer 4

The first thing I notice is that
x²+ 7x + 12 factorises as (x+3)(x+4)
and
x²-9 factorises as (x+3)(x-3)

Hence
log(x²-9) - log(x²+7x+12)
= log[(x+3)(x-3)]-log[(x+3)(x+4)]
= log[(x+3)(x-3)] / [(x+3)(x+4)]
= log [ (x-3) / (x+4) ]

which is just about as tidy as you'll get it.

Answer 5

log ( (x^2-9) / ( x^2+7x+12 ) ) = log ( (x-3)/(x+4) )

Answer 6

log [ (x^2-9) / (x^2 = 7x + 12) ]

on the basis that log x - log y = log (x/y)

Answer 7

If log10y+3log10x=2, express y in terms x

Answer 8

= log [ (x - 3).(x + 3) / (x + 3).(x + 4) ]
= log [ (x - 3) / (x + 4) ]

Answer 9

i hate log