How do you solve #2 log x^4 = 16#?

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Answer 1 · 2015-02-19T15:58:59

It depends upon the base of your logarithm. Let us assume that the base is a number #a#.

You have:

#2log_ax^4=16#

#log_ax^4=16/2# the power of #4# can go as multiplier of log:

#4log_ax=8# and again:

#log_ax=8/4#

#log_ax=2#

#x=a^2#

Now, you can choose the value of #a#. Normally (when it is not specified) it should be #10#, so if this is the case you get:
#x=10^2=100#