How do I solve a log base 2 equation?

log₂(x + 3) + log₂(x - 3) = 4

If you have the sum of two logs, it's the same as the log of the products:

log(x) + log(y) = log(xy)

So we can combine the two logs into one:

log₂[(x + 3)(x - 3)] = 4

Now that we're here, if we make both sides an exponent over a base of 2, that undoes the log and leaves only what's inside. Now we have:

(x + 3)(x - 3) = 2⁴

Now we can simplify both sides:

x² - 9 = 16

Add 9 to both sides:

x² = 25

and square root of both sides:

x = ±5

Note that you cannot have the log of a negative number, so we can throw out the -5 to leave this as your only answer:

x = 5

log₂(x+3) + log₂(x-3) = 4

log₂(x+3)(x-3) = 4

(x+3)(x-3) = 2⁴

x² - 9 = 16

x² = 25

x = 5

(Note other answer "-5" makes the numbers (x+3) and (x-3) as -ve. Since log of -ve number is not allowed, we ignore this answer)

Combine the terms on the left into one term. Write the term on the right as log base 2 of something.

log_2 [ one big term ] = log_2 [2^4]

Then [one big term] = 2^4 = 16.

Solve that for x.