Question #3c7cf

1 Answer
Jun 16, 2017

See the explanation below.

Explanation:

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