How do you solve #\log _ { 6} x + \log _ { 6} ( x + 1) = 1#?

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Answer 1 · 2017-01-01T14:55:44

#x=2#

using the law of logs:

#log_aX+log_aY=log_aXY#

#log_6x+log_6(x+1)=1#

#log_6x(x+1)=1#

using the definition #log_ab=c=>a^c=b#

the RHS becomes #log_(6)6#

ie. #" "log_6x(x+1)=log_(6)6#

#=>x(x+1)=6#

simplifies to:

#x^2+x-6=0#

factorising

#(x+3)(x-2)=0#

solving: #" " x=-3" "# or#" "x=2#

but #log_aX" " #is only defined for #X>0#

#:. x=2#