Question

How do you solve `log_4(x) + log_4(x+2) = 1/2 log_4 9`?

Answer 1

`x=1` and `x=-3`

Explanation:

Start from the given equation:

`log_4 x + log_4 (x+2)= 1/2 log_4 9`

`log_4 x(x+2)= log_4 9^(1/2)`

`Antilog(log_4 x(x+2))= Antilog( log_4 9^(1/2))`

`x(x+2)= 9^(1/2)`

`x(x+2)= 3`

`x^2+2x-3=0`

`(x+3)(x-1)=0`

`x+3=0` and `x-1=0`

`x=-3` and `x=1`

Have a nice day from the Philiipines !!!

Answer 2

x = 1

Explanation:

Using the following 'laws of logs'

`• logx + logy = logxy`

`• logx^n = nlogx`

then `log_4 x + log_4(x+2) = log_4 9^(1/2)`

hence `x ( x+ 2 ) = 9^(1/2) = 3`

( since if `log_b a = log_b c color(black)(" then ") a = c )`

so x (x+ 2 ) = 3 → `x^2 + 2x - 3 = 0`

Factors to give : (x + 3 )(x - 1 ) = 0

`rArr x = - 3 , x = 1`

but x ≠ - 3 hence x = 1

Answer 3

`x=1` and `x=-3`

Explanation:

For `x=-3` is also a solution considering that
the absolute value of x and (x+2) in the equation may be used.

example:

`log_4 abs(x)+log_4 abs(x+2)=1/2*log_4 9`

`log_4 abs(-3)+log_4 abs(-3+2)=1/2*log_4 9`

`log_4 3+log_4 1=1/2*log_4 9`

`log_4 3+ 0=log_4 9^(1/2)`

`log_4 3=log_4 3`

therefore `x=-3` is a solution also !