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Solve the value of x; 2+(㏒x)^2=3 ㏒x

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Answer:

x = 10 or x = 100

Explanation:

The domain of the equation: x > 0.

We have:


2+(\log x)^2=3\log x

Substitute
\log x=t:


2+t^2=3t subtract 3t from both sides


t^2-3t+2=0


t^2-2t-t+2=0


t(t-2)-1(t-2)=0


(t-2)(t-1)=0\iff t-2=0\ \vee\ t-1=0


t-2=0 add 2 to both sides


t=2


t-1=0 add 1 to both sides


t=1

We return to substitution:


t=2\to\log x=2

and


t=1\to\log x=1

Use


\log_ab=c\iff a^c=b


\log a=\log_(10)a


\log_aa=1


\log x=2\iff x=10^2\to x=100\\\\\log x=1\to x=10

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