solving logs?

Question

log(5) (x-1) + log(5) (x-2) - log(5) (x+6) = 0

**(5) means to the base of 5
** please help step by step
*** the awnser is suposed to be
x=2 + 2 [square root of 2]

Answer 1

log [ (x - 1)(x - 2) / (x + 6) ] = 0
(x - 1)(x - 2) / (x + 6) = 5^0 = 1
x² - 3x + 2 = x + 6
x² - 4x - 4 = 0
x = [ 4 ± √(16 + 16)] / 2
x = [ 4 ± √(32) ] / 2
x = [ 4 ± 4√(2) ] / 2
x = [ 2 ± 2√(2) ]

Answer 2

Remember that log base 5 of (x-1) is the exponent on 5 that gives x-1.
Notice that 5^[log(5)(a)+log(5)(b)-log(5)(c)]
=(a*b)/c and remmeber 5^0=1
This gives
(x-1)(x-2)/(x+6)=1 So
(x-1)(x-2)=(x+6) and
xx-3x+2=x+6 So
xx-4x-4=0
Now use quadratic formula
a=1, b= -4 c=-4
x= -(-4)/2 +/-sqrt[(-4)(-4)-4(1)(-4)]/2
=2+/-sqrt[16+16]/2
=2+/-sqrt[32]/2
=2+/-sqrt[32/4]
=2+/-sqrt(8)
=2+/-sqrt(2*4)
=2+/- 2*sqrt(2)
Since x has to be positive, discard
the negatve solution

Answer 3

by the laws of logarithms
adding logs means multiply, subtracting means divide
so log(5) (x-1) + log(5) (x-2) - log(5) (x+6)
= log(5) [(x-1)(x-2)/(x+6)]

make that equal 0
log(5) [(x-1)(x-2)/(x+6)] = 0

put both sides as powers of 5
5^(log(5) [(x-1)(x-2)/(x+6)]) = 5^0
(x-1)(x-2)/(x+6) = 1

multiply by (x+6)
(x-1)(x-2) = x+6

multiply out
x² - 3x + 2 = x + 6
x² - 4x -4 = 0

complete the square
(x - 2)² -8 = 0
(x - 2)² = 8
x - 2 = ±√8
x = 2±√8
x = 2±2√2

since x>2
x = 2+2√2

Answer 4

Divide everything by log 5. Take the inverse log to get
(x-1)(x-2)/(x+6)=1

Multiply both sides by (x+6) to get a quadratic equation which you can solve using the formula.

Answer 5

log(5) (x-1) + log(5) (x-2) = log(5) (x+6)
Since you have the same base, you get
(x-1)(x-2) = x+6, x > 2 (required by log(5)(x-2)...)
x^2-4x+2 = 0
Use quadratic formula to solve for x,
x = 2 + √2
This is the only solution since x > 2.

Answer 6

log(5) (x-1) + log(5) (x-2) - log(5) (x+6) = 0
log(5)[(x-1)(x-2) / (x+6)] = 0
(x-1)(x-2) / (x+6) = (5) ^ 0
(x-1)(x-2) / (x+6) = 1
(x-1)(x-2) = x + 6
x ^ 2 -2 x -x +2 = x + 6
x ^ 2 -4x -4 = 0 solve yourself .armenian Star...

Answer 7

ok calm down

in the first part of the equation you are adding two logs of the same base (5).

-take these two quantities (x-1) and (x-2) and multiply them
log (5) (x-1)(x-2) - log (5) (x+6) = 0

now you have two logs of the same base subtracting each other, move one of the logs two the other side of the equation by adding it. soo...
log (5) (x-1) (x-2) = log (5) (x+6)

now you have two logs of the same base equal to each other. Simply take their quantities and make them equal to each other now, ignoring the logs

(x-1) (x-2) = (x+6)

now multiply out of your qrouping symbols and I'm sure, you'll get the answer you're looking for

hope that helped!

Answer 8

while achievable, use residences... thus it fairly is achievable. the valuables you ought to use is: The log of potential of the backside is the exponent... it fairly is... in base 4 log(4^7) = 7 First you write (a million/sixteen) as potential of four.... (a million/sixteen)^3 = (4^-2) And now prepare the valuables... (base 4) [log4^(-2)]^3 = [-2]^3 = -8 ok!

Answer 9

what does log supposed to mean?

Answer 10

nah