Question

log base 6 (x-1) + log base 6 (x+4)=2

Answer 1

this means 6^2 = (x-1)(x+4)
36 = x^2+3x-4
x^2+3x-40 =0
x= 5 or -8

Answer 2

careful!

Answer 3

there is a small error here.

Answer 4

ya thats what i thought..

Answer 5

Nick Black Champion 0
log6(x−1)+log6(x+4)=2

log6(x−1)(x+4)=2

x2+3x−40=0

(x+8)(x−5)=0

x=-8,5

Answer 6

you have to use anti logs

Answer 7

where di u get 40?

Answer 8

\[\log_6((x-1)(x+4))=2\]
\[6^2=(x-1)(x+4)\]
\[x^2+3x-4=36\]
\[x^2+3x-40=0\]
\[(x+8)(x-5)=0\]

Answer 9

where is the error

Answer 10

all this is good, but
\[x=-8,x=5\] is wrong because you cannot take the log of a negative number so answer is only
\[x=5\]

Answer 11

thanks for the info

Answer 12

no prolbem

Answer 13

Yeah anti logarithms are greater than 0, so you have
\[x - 1 > 0 , x+4>0\]
Which results to \[x >1\]
Therefore the solution x = -8 does not satisfy this condition.
So x = 5 As satellite73 put.