Question

How do you solve `Log_2x + log_2(x+2) = log_2(6x+1)`?

Answer 1

`x_1=2+sqrt(5) or x_2=2-sqrt(5)`

Explanation:

`Log_2x + log_2(x+2) = log_2(6x+1)`

Remember that `log(a)+log(b)=log(a*b)`

`Log_2(x*(x+2)) = log_2(6x+1)`
`x*(x+2)=6x+1`
`x^2+2x=6x+1|-2x`
`x^2=4x+1|-4x-1`
`x^2-4x-1=0`
`(x-2)^2-1-4=0|+5`
`(x-2)^2=5|sqrt()|+2`
`x=2+-sqrt(5)`
`x_1=2+sqrt(5) or x_2=2-sqrt(5)`
`x_2<0 -> log(x_2)∈CC`

Answer 2

`x=2+sqrt5`

Explanation:

`log_2x+log_2(x+2)=log_2(6x+1)`

`color(green)(loga+logb=logab`

`log_2x(x+2)=log_2(6x+1)`

By taking both sides for powers of 2 like this:

`2^(log_2x(x+2))=2^(log_2(6x+1)`

`color(green)(a^(log_ax)=x`

Thus

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

`x^2+2x=6x+1`

`x^2-4x-1=0`

`x=2+sqrt5`

`" OR "`

`x=2-sqrt5` `color(red)" refused as it doesn't satisfy the equation"`

Answer 3

`color(blue)(x=2+sqrt(5))`

Explanation:

By the laws of logarithms:

`log_a(b)+log_a(c)=log_a(bc)`

`log_a(b)=log_a(c)<=>b=c`

# #

`log_2(x)+log_2(x+2)=log_2(6x+1)`

`log_2(x(x+2))=log_2(6x+1)`

`log_2(x^2+2x)=log_2(6x+1)`

`:.`

`x^2+2x=6x+1`

`x^2-4x-1=0`

By quadratic formula:

`x=(4+-sqrt(16+4))/2=(4+-2sqrt(5))/2=2+-sqrt(5)`

We need to test this:

`x=2-sqrt(5)<0`

For:

`log_2(x)` This is undefined for real mumbers.

So only the soloution:

`color(blue)(x=2+sqrt(5))`