How do you solve #2log_10 6 - 1/3 log_10 27 - log_10 x = 0#?

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Answer 1 · 2016-03-31T23:03:29

First, use the log rule #alogn = loga^n#

#log_10(6^2) - log_10(27^(1/3)) - log_10(x) = 0#

Evaluate, using the exponent rule #a^(n/m) = root(m)(a^n)#

#log_10(36) - log_10(3) - log_10(x) = 0#

Now, use the log rule #log_am - log_an = log_a(m/n)#

#log_10((36/3)/x) = 0#

#log_10(12/x) = 0#

Convert to exponential form: #log_an = x -> a^x = n#

#10^0 = 12/x#

#1 = 12/x#

#x = 12#

Hopefully this helps!