How do you solve #log (x²+7)=2/3 log 64#?

1 Answer
Noah G
Mar 15, 2016

You first of all must put all logs to one side of the equation.

Explanation:

#log(x^2 + 7) - 2/3log64 = 0#

Simplify using the rule #alogn = logn^a#

#log(x^2 + 7) - log64^(2/3) = 0#

#log(x^2 + 7) - logroot(3)(64^2) = 0#

#log(x^2 + 7) - log16 = 0#

Simplify using the rule #log_an - log_am = log_a(n/m)#

#log((x^2 + 7)/(16)) = 0#

Convert to exponential form #(log_bn = x => b^x = n)#. The log is in base 10, since nothing is noted in subscript beside the log.

#(x^2 + 7)/16 = 10^0#

Simplify by the property #a/b = m/n -> a xx n = b xx m#

#x^2 + 7 = 1(16)#

#x^2 = 16 - 7#

#x^2 = 9#

#x = +-3#

Normally, in logarithmic equations, a negative solution must be rejected because if a log is negative it is undefined. Therefore, it is always extremely important to verify your solutions inside the original equation. However, in this case both solutions are fine so your solution set is #{x = +-3}#.

Hopefully this helps!

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