How does log2 (x+1)+3 equal 8(x+1)?

[linapril]

Hey! Could someone explain how log₂ (x+1)+3 equals 8(x+1)? Thank you!

 

[Nono713]

It doesn't, but I think it's a typo. I'll go on this assumption.

First look at $8(x + 1)$. Take its base 2 logarithm:

$$\log_2{(8(x + 1))}$$

Now remember the logarithm product rule:

$$\log{(xy)} = \log{(x)} + \log{(y)}$$

So apply this to what you have right now:

$$\log_2{(8(x + 1))} = \log_2{(8)} + \log_2{(x + 1)}$$

And since $2^3 = 8$, $\log_2{(8)} = 3$, and you can conclude:

$$\log_2{(8(x + 1))} = \log_2{(x + 1)} + 3$$