Question

How do you solve `log (x+4)=log x + log 4`?

Answer 1

Put all logs to one side of the equation and solve using the property `log_an - log_am = log_a(n/m)`

Explanation:

`log(x + 4) - logx = log4`

`log(((x + 4)/(x))/4 )= 0`

Convert to exponential form. The base is 10, since nothing is noted in subscript in the log.

`(((x + 4)/x) / 4) = 10^0`

`(x + 4)/x = 1 xx4`

x + 4 = 4x

4 = 3x

`4/3` = x

Hopefully this helps.

Answer 2

`x=4/3`

Explanation:

`log(x+4)-log(x)=log(4)`

`log((x+4)/x)=log(4)`

`log((x+4)/x)-log(4)=0`

`log((x+4)/x xx1/4)=0`

`log((x+4)/(4x))=0`

Consider standard form: `log_10(x)=y -> 10^y=x`

so`" " log_10((x+4)/(4x))=0 -> 10^0=(x+4)/(4x)`

Giving: `" "4x=x+4`

`3x=4`

`x=4/3`