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!