How do you solve \log _ { 9} x ^ { 2} = 5log9x2=5?

2 Answers
Dec 7, 2017

x=243 or x=-243x=243orx=243

Explanation:

log_9x^2=5log9x2=5

log_9x^2=log(x^2)/log9log9x2=log(x2)log9

log(x^2)/log9=5log(x2)log9=5

Multiply both sides by log9

log(x^2)=5log9log(x2)=5log9

5log9=log(9^5)=log(59049)5log9=log(95)=log(59049)

logx^2=log(59049)logx2=log(59049)

Cancel logarithms by taking exp of both sides

x^2=59049x2=59049

take square root of both sides

x=243 or x=-243x=243orx=243

Dec 7, 2017

x=+-3^5=+-243x=±35=±243

Explanation:

we can use the definition of logs

log_ab=c=>a^c=blogab=cac=b

log_9x^2=5=>9^5=x^2log9x2=595=x2

x^2=9^5=(3^2)^5x2=95=(32)5

=>x==+-sqrt((3^2)^5)=+-(sqrt(3^2))^5x==±(32)5=±(32)5

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