Question

Note:Base 4= B4
logB4(x+3)+logB4(x-3)=2

Answer 1

step is to convert it to
log4[(x-3)(x+3)] = 2

Now we raise 4 to the power of both sides
(x-3)(x+3) = 4^2

Simplify
x^2 - 9 = 16

Add 9 to both sides
x^2 = 25

x = 5 or x = -5

Answer 2

Shay, I'm so sorry about @Raymond67 's totally unacceptable comment.

Answer 3

@LegoMyEgo so you change it to exponential form?

Answer 4

@mathmale Thanks for removing it.

Answer 5

Several things:

First, it's true that logB4(x+3)+logB4(x-3) = logB4 [(x+3)(x-3)]. This stems from one of the most important rules of logs.

Answer 6

how do you know when to change it?

Answer 7

Secondly, y=logB4 x and y=4^x are INVERSE FUNCTIONS.

Answer 8

Are you, Shay, familiar with this term, inverse function?

Answer 9

Shay, do you wan Mathmale to explain it to you?

Answer 10

yes

Answer 11

kk

Answer 12

Shay:
Rule 1: log a + log b = log a*b (multiplication rule)
Rule 2: log a - log b = log (a/b) (division rule)
Rule 3: log a^b = b*log a (power rule)

I'd be happy to discuss these with you further if you'd like.

But back to the original logarithmic equation:

Answer 13

Using Rule 1: logB4 (x+3) + logB4 (x-3) = logB4 (x+3)(x-3).
Before we move on, are you OK with this?

Answer 14

yes @ mathmale