How do you solve $log_2(x-4)+log_2(x+4)=3$ ?

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Answer 1 · 2016-03-09T02:41:20

You must use the log property $log_am + log_an = log_a(m xx n)$

Note: $log_am - log_an = log_a(m/n)$

$log_2(x - 4)(x + 4) = 3$

Convert to exponential form.

$x^2 - 16 = 2^3$

$x^2 - 24 = 0$

$x^2 = 24$

$x = sqrt(24) = 2sqrt(6)$

Note: since the log of a negative number is non-defined, we cannot have $x = +-sqrt(24)$ , we can only have $+sqrt(24)$

Practice exercises:

a) $log_3(2x + 1) + log_3(3x - 4) = 6$

b) $log_4(5x - 2) - log_4(3x + 1) = 2$

Challenge problem:

Solve for x in the following equation.

$y = log_2(4x + 1) + log_2(x + 3)$

Hopefully this helps!

How do you solve log_2(x-4)+log_2(x+4)=3log_2(x-4)+log_2(x+4)=3?