(logx)^2+logx^5=-6 (Hint: (logx)^2≠2logx)?

Answer 1

(logx)^2+log(x^5)=-6
(logx)^2+5logx+6=0
let y = logx
y^2 + 5y + 6 = 0
(y+3)(y+2)=0
y = -2 or -3
logx = -2 or -3
x = 10^-2 or 10^-3
x = 1/100 or 1/1000

Answer 2

(logx)^2+logx^5=-6
let logx=t
then t =-2 or -3.so,x=1/100 or 1/1000;(if base is 10,otherwise if base is 2 then the answers are 1/4 or 1/8)

Answer 3

(logx)^2+logx^5=-6
So (logx)^2 + 5 logx + 6 = 0
(logx + 2)(log x + 3) = 0
log x = -3, -2
x = 0.01, 0.001

Answer 4

(logx)^2+logx^5= -6
or (logx)^2+5logx + 6= 0
or (logx +2 )(logx + 3) = 0
or, logx = -2 or logx = -3
or x = 10^-2 or x = 10^-3
or, x= 0.01 or x = 0.001

Answer 5

Let u = log(x); the equation can then be written

u^2 + 5*u + 6 = 0

(u + 2)*(u + 3) = 0

u = -2 and u = -3

log(x) = -2 and log(x) = -3;

x = 10^-2 , x = 10^-3

Answer 6

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Answer 7

The question should be corrected in the second term of LHS as (log x)^5

Answer 8

what's that?