Question

log5(3x+1)+log5(x+1)=1
the (log5), 5 is a bit down the log :)

Answer 1

(3x+1)(x+1)/5

Answer 2

start with
\[\log_2((3x+1)(x+1))=1\] then
\[(3x+1)(x+1)=5^1\]

Answer 3

1 is log5(5)

Answer 4

than coming on the left part will be with sign minus than result divide

Answer 5

then
\[3 x^2+4 x-4 = 0\] and then
\[(x+2)(3x-2)=0\] so
\[x=\frac{2}{3}\] and you can ignore the negative answer because you cannot take the log of a negative number

Answer 6

satellite why not is right ???

Answer 7

@krypot
\[\log_5(x)=y\iff x = 5^y\]

Answer 8

*krypton, sorry

Answer 9

lol

Answer 10

so
\[\log_5((3x+1)(x+1))=1\iff (3x+1)(x+1)=5\]

Answer 11

am getting 2 as my answer

Answer 12

yo am getting *2* as my answer

Answer 13

got -2

Answer 14

3x2 +4x+1=0
x',2=(-4+/- sqrt(16-12))/6 =(-4+/- 2)/6 = -1 and - 1/3

Answer 15

how you have got this result ?

Answer 16

i did log5(3x+1)/(x+1)=5
then cross multiplied it

Answer 17

i got (3x+1)/(x+1)=5
i then cross mulitplied

Answer 18

no when we have loga + logb=log(a*b)

Answer 19

you need multiply not divide

Answer 20

oh! thanks almost forgot my laws of indices.lol

Answer 21

bye and good luck