Question

If 4log10⁡[x+1]−6log10⁡x−2⋅3log10⁡[x2+2]=0 then the value of 120x is

Answer 1

Explanation:
Consider,4log10⁡(x)+1−6log10⁡x−2⋅3log10⁡(x2)+2=0⇒[22]log10⁡(x)+1−[2⋅3]log10⁡x−2⋅3log10⁡x2⋅32=0⇒22log10⁡x+2−2log10⁡x⋅3log10⁡x−2⋅3log10⁡x2⋅32=0[Using properties of logarithmic function-3 ]⇒4⋅2log10⁡x2−2log10⁡x⋅3log10⁡x−18⋅32log10⁡x=0Now, substitute 2log10⁡x=A and 3log10⁡x=B, we get4A2+AB−18B2=0 [Using middle term split ]⇒(4A−9B)(A+2B)=0But, A+2B≠0 as A>0 AND B>0⇒4A=9B⇒4⋅2log10⁡x=9⋅3log10⁡x⇒49=(32)log10⁡x⇒(32)−2=(32)log10⁡x⇒log10⁡x=−2[Using properties of logarithmic function-10]⇒x=1100Thus, 120x=5Hence, the answer is 5.00.