How do you solve #\log _ { 9} x ^ { 2} = 5#?

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Answer 1 · 2017-12-07T03:43:16

#x=243 or x=-243#

#log_9x^2=5#

#log_9x^2=log(x^2)/log9#

#log(x^2)/log9=5#

Multiply both sides by log9

#log(x^2)=5log9#

#5log9=log(9^5)=log(59049)#

#logx^2=log(59049)#

Cancel logarithms by taking exp of both sides

#x^2=59049#

take square root of both sides

#x=243 or x=-243#

Answer 2 · 2017-12-07T08:48:52

#x=+-3^5=+-243#

we can use the definition of logs

#log_ab=c=>a^c=b#

#log_9x^2=5=>9^5=x^2#

#x^2=9^5=(3^2)^5#

#=>x==+-sqrt((3^2)^5)=+-(sqrt(3^2))^5#

#:.x=+-3^5=+-243#