How do you solve #\log _ { 8} ( x - 6) + \log _ { 8} ( x + 6) = 2#?

1 Answer
Noah G
Dec 18, 2016

#x = 10#.

Explanation:

Use the sum formula of logarithms #color(magenta)(log_a n + log_a m = log_a(n xx m)# to express as a single logarithm.

#=>log_8[(x- 6)(x + 6)] = 2#

Convert to exponential form using #color(magenta)(log_a n = b -> a^b = n#.

#=> x^2 - 36 = 8^2#

#=>x^2 - 36 = 64#

#=> x^2 - 36 - 64= 0#

#=> x^2 - 100 = 0#

#=> x^2 = 100#

#=> x = +- 10#

However, the domain of #y = log_8 x# is #x > 0#. Similarly, the domain of #y = log_8 (x + 6)# is #x > -6# and that of #y = log_8 (x - 6)# is #x > 6#. To satisfy the equation, therefore, #x > 6#, otherwise #log_8(x - 6)# will become undefined.

Hence, #color(green)(x = 10)# is the only valid solution; #color(red)(x = -10)#, though, is extraneous.

Hopefully this helps!

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