Question

3^ log base 3 of log √x - log x - log^x - 3 = 0

Answer 1

Answer:

✌❤

Step-by-step explanation:

Using property of log: alog(base"a")(x)=x

So, (3(log(√x−1)"to"base"3)=sqrt(x−1)

Similarly, (3log(x−6)"to"base"3)=x−6

So, sqrt(x−1)<x−6+3

sqrt(x−1)<x−3

x−1<x2−6x+9

x2−7x+10>0

x2−5x−2x+10>0

(x−5)(x−2)>0

So, x>5 and x<2

But x>1 due to sqrt(x−1) term in the equation…

Also x>6 due to log(x−6) term in the equation…

So, the final answer should be x>6

Hope this was helpful..

Regards

Answer 2

Answer:

Step-by-step explanation:

3log3logx√−logx+(logx)2−3=03log3logx-logx+(logx)2-3=0  

We know, alogab=balogab=b, so our equation becomes,

⇒logx−−√−logx+(logx)2−3=0⇒logx-logx+(logx)2-3=0

⇒12logx−logx+(logx)2−3=0⇒12logx-logx+(logx)2-3=0

⇒(logx)2−logx2−3=0⇒(logx)2-logx2-3=0

⇒2(logx)2−logx−6=0⇒2(logx)2-logx-6=0

⇒2(logx)2−4logx+3logx−6=0⇒2(logx)2-4logx+3logx-6=0

⇒(2logx+3)(logx−2)=0⇒(2logx+3)(logx-2)=0

⇒logx=−32andlogx=2⇒logx=-32andlogx=2

⇒x=e−32andx=e2.