How do you solve #6log_3(0.5x)=11#?

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Answer 1 · 2017-05-04T05:57:48

#x=2xx3^(11/6)~=2xx7.5~=15#

Let's approach it by first dividing through by 6:

#6log_3(0.5x)=11#

#log_3(0.5x)=11/6#

We can now make both sides the exponent of a base 3 (which will be the inverse function for the left hand side and cancel out the log):

#3^(log_3(0.5x))=3^(11/6)#

#0.5x=3^(11/6)#

#x=2xx3^(11/6)~=2xx7.5~=15#