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Answer 1 · 2017-06-16T04:47:01

See the explanation below.

I'm not sure if this is the exact question, I added a minus sign. But this is a similar question.

We will apply the following log law twice to solve for x:

#log_a(b)-log_a(c)=log_a(b/c)#

First, see if you can simplify large numbers and write them in index form:

#log_3x-log_3(x/81)=log_3(x/729)#

#rArrlog_3x-log_3(x/3^4)=log_3(x/3^6)#

Apply the log law to the left hand side:

#rArrlog_3(cancel(x)/(cancel(x)/3^4))=log_3(x/3^6)#

#rArrlog_3(3^4)=log_3(x/3^6)#

Bring both terms to the same side and apply the log law again:

#rArrlog_3(3^4)-log_3(x/3^6)=0#

#rArrlog_3((3^4)/(x/3^6))=log_3((3^4*3^6)/x)=log_3((3^10)/x)=0#

Write the log expression in it's equivalent exponential form and then solve for x:

#log_3((3^10)/x)=0rArr3^0=(3^10)/x=1#

#rArrx=3^10#