5
Question
Solve the following equation.
3+2logx+13=2log3(x+1)
3+2logx+13=2log3(x+1)
Open in App
Solution
3+2logx+13=2log3x+1
Let log3x+1=a
Therefore, 3+2a=2a2a2−3a−2=0⇒2a2−4a+a−2=0⇒(2a+1)(a−2)=0⇒a=−12,2
Therefore, log3x+1=−12,log3x+1=2⇒x+1=1√3,9⇒x=1−√3√3,8.
3+2logx+13=2log3x+1
Let log3x+1=a
Therefore, 3+2a=2a2a2−3a−2=0⇒2a2−4a+a−2=0⇒(2a+1)(a−2)=0⇒a=−12,2
Therefore, log3x+1=−12,log3x+1=2⇒x+1=1√3,9⇒x=1−√3√3,8.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
