Question

What is x if `log_4 x=2 - log_4 (x+6)`?

Answer 1

See process below

Explanation:

In this type of equations, our goal is to arrive to an expresion like
`log_bA=log_bC` from this, we conclude `A=C`

Lets see

`log_4 x=2-log_4(x+6)`

We know that `2=log_4 16`. then

`log_4 x=log_4 16-log_4(x+6)=log_4(16/(x+6))` applying the rule

`log(A/B)=logA-logB`

So, we have `x=16/(x+6)`

`x^2+6x-16=0`
by quadratic formula

`x=(-6+-sqrt(36+64))/2=(-6+-10)/2`

Solutions are `x_1=-8` and `x_2=2` we reject negative and the only valid solution is `x_2`

Answer 2

`x=2`

Explanation:

`"using the "color(blue)"laws of logarithms"`

`•color(white)(x)logx+logy=log(xy)`

`•color(white)(x)log_b x=nhArrx=b^n`

`" add "log_4(x+6)" to both sides"`

`log_4x+log_4(x+6)=2`

`log_4x(x+6)=2`

`x(x+6)=4^2=16`

`x^2+6x-16=0`

`(x+8)(x-2)=0`

`x=-8" or "x=2`

`x>0" and "x+6>0`

`"thus "x=-8" is invalid"`

`rArrx=2" is the solution"`