How do you solve #\log _ { 27} ( 2x + 4) + \log _ { 81} ( 6x + 12) = \frac { 5} { 6}#?

1 Answer
Mar 8, 2018

Answer:

#x=-1/2#

Explanation:

Using #logab=logb/loga#, we can simplify given equation and solve it as follows:

#log_27(2x+4)+log_81(6x+12)=5/6#

i.e. #log(2x+4)/log27+log(6x+12)/log81=5/6#

or #log(2x+4)/(3log3)+log(6x+12)/(4log3)=5/6#

or #log(2x+4)/3+log(6x+12)/4=5/6log3#

or #4log(2x+4)+3log(6x+12)=10log3#

or #log[(2x+4)^4(6x+12)^3]=log3^10#

or #(2x+4)^4(6x+12)^3=3^10#

or #2^4(x+2)^4*6^3(x+2)^3=3^10#

or #(x+2)^7=3^10/(2^4*6^3)#

or #(x+2)^7=3^10/(2^4*2^3*3^3)#

or #(x+2)^7=3^7/2^7#

or #x+2=3/2#

or #x=3/2-2=-1/2#