How do you solve #4log_4 (x) - 9log_x (4) = 0#?

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Answer 1 · 2016-03-14T15:08:13

You must first simplify using the rule #log_am = logm/loga#

First, we must use the rule #alogn = logn^a#

#log_4(x^4) - log_x(262144) = 0#

#(logx^4)/(log4) - (log262144)/logx = 0#

Place on a common denominator.

#(logx(logx^4))/(log4(logx)) - (log4(log262144))/(logx(log4)) = 0#

Convert to exponential form.

#x^5/1048576 = 10^0#

#x^5 = 1048576#

#x = root(5)1048576#

#x = 16#

Hopefully this helps!