#log_3(x+12)+log_3(x-12)=4# ?

2 Answers

Answer:

#x=15#

Explanation:

We have:

#log_3(x+12)+log_3(x-12)=4#

We can combine the left side by saying:

#log_3((x+12)(x-12))=4#

Know that we can rewrite log questions using the following format:

#a^b=c <=> log_(a)c=b#

So we can write out question as:

#3^4=(x+12)(x-12)#

And let's expand:

#81=x^2-144#

#x^2=225#

#x=+-15#

And now let's test the solutions:

For #x=15#

#log_3(x+12)+log_3(x-12)=4#

#log_3(15+12)+log_3(15-12)=4#

#log_3(27)+log_3(3)=4#

#3+1=4#

For #x=-15#

#log_3(-15+12)+log_3(-15-12)=4#

#log_3(-3)+log_3(-27)=4# and this doesn't work because we can't have negatives inside the log function.

Therefore our solution is #x=15#

Oct 11, 2016

Answer:

#x = 15#

Explanation:

I have assumed that you meant #log_3# as I have now shown in the question.

First problem:
You can't mix log terms and number terms, all one or all the other.

Second problem:
How do you write 4 as #log_3#?

Compare:
#10^1 = 10 hArr log_10 10 = 1" and "3^1 =3 hArr log_3 3=#
#10^2 = 100 hArr log_100 = 2" and " 3^2 = 9 hArr log_3 9 = 2#
#10^3 = 1000 hArr log_1000 = 3" and "3^3 = 27 hArr log_3 27 =3 #

#log_3 (x+12) + log_3 (x-12) = color(red)(4)" "larr# can be written as:
#log_3 (x+12) + log_3 (x-12) = color(red)(log_3 81)#

Condense the two log terms into one:

#log_3 (x+12)(x-12) = log_3 81#

Now use the log law: If #log A = log B hArr A = B#

#cancel(log_3) (x+12)(x-12) = cancel(log_3) 81#

#(x+12)(x-12) = 81" "larr# notice the difference of squares

#x^2 -144 = 81#

#x^2 = 225#

#x = +-sqrt225 = +-15#

However in this question, #-15# is not valid.

#x = 15#

Be sure to notice the base is 3, not 10!