How to solve the equation log_(x) 5+log_(5) x=5/2logx5+log5x=52 ?

1 Answer
Jan 25, 2018

x_1=sqrt5x1=5 and x_2=25x2=25

Explanation:

log_x5+log_5x=5/2logx5+log5x=52

log_x5+1/log_x5=5/2logx5+1logx5=52

After using y=log_x5y=logx5, this equation became

y+1/y=5/2y+1y=52

(y^2+1)/y=5/2y2+1y=52

2*(y^2+1)=5*y2(y2+1)=5y

2y^2+2=5y2y2+2=5y

2y^2-5y+2=02y25y+2=0

(2y-1)*(y-2)=0(2y1)(y2)=0

From this equation, y_1=1/2y1=12 and y_2=2y2=2

For y=2y=2, log_x5=2logx5=2 or x^2=5x2=5. So x_1=sqrt5x1=5

For y=1/2y=12, log_x5=1/2logx5=12 or x^(1/2)=5x12=5. So x_2=5^2=25x2=52=25