Show that #1/2log9+2log6+1/4log81-log12=3log3#?

3 Answers
George C.
May 16, 2018

See explanation...

Explanation:

If #a, b > 0# then:

#log ab = log a + log b#

and

#log (a/b) = log a - log b#

Hence:

#log a^n = n log a" "# for any integer #n > 0#

So we find:

#1/2 log 9 + 2 log 6 + 1/4 log 81 - log 12#

#=1/2 log 3^2 + 1/4 log 3^4 + log 6^2 - log 12#

#=1/2 (2 log 3) + 1/4 (4 log 3) + log (36/12)#

#= log 3 + log 3 + log 3#

#= 3 log 3#

Jim G.
May 16, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx^nhArrnlogx#

#•color(white)(x)logx+logyhArrlog(xy)#

#•color(white)(x)logx-logy=log(x/y)#

#"consider left side"#

#=log9^(1/2)+log6^2+log81^(1/4)-log12#

#=log3+log36+log3-log12#

#=log((3xx36xx3)/12)#

#=log27#

#=log3^3=3log3#

Bhavin B.
May 16, 2018

Please see the explanation below in the explanation section.

Explanation:

Be familiar with some Rules of Logarithm:

#log_a MN# = #log_a M + log_a N#

#log_a(M/N)# = # log_a M − log_a N#

#log_aM^k# = #k log_aM#

Now lets see the original equation:

#1/2log9+2log6+1/4log81-log12=3log3#

Step 1:
Lets simplify each term now.

First term:

#color(blue)(1/2log9)# = #1/2log3^2# = #2xx1/2log3# = #color(red)log3#

Second Term

#color(blue)(2log6)# = #log6^2# = #log36# = #log(4x9) = #log 4 + log 9# = log 4 + log3^2# = #color(red)log4 + color(red)(2log3)#

Third Term

#color(blue)(1/4(log81))# = #(1/4)log3^4# = #4 xx(1/4)log3# = #color(red)log3#

Fourth Term

#color(blue)log12#= #log (3 xx4)# =# color(red)log 3 + color(red)log 4#

Step 2:
Add all Terms:

#log 3 + log 4 + 2 log 3+ log 3 - (log3 + log 4)#

#log 3 + log 4 + 2 log 3 + log 3 - log 3 - log 4#

#log 3 + log 4 - log 4 + 2 log 3 + log 3 - log 3 #

#log 3 + cancel (log 4 - log 4) + 2 log 3 + cancel(log 3 - log 3) #

#log 3 + 2 log 3#
#log 3 +2 log 3# = #log 3 + log3^2# = log 3 + log 9# = #log (3 xx 9)# = ##log 27# = #log 3^3# = #color(red)(3log3)#

Hence, we have proved that:
#1/2log9+2log6+1/4log81-log12=3log3# = #3log3#

Hope all steps are understood. Cheers!

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