Answer:
x = 3⁸
Step-by-step explanation:
Step(i):-
Given that
[tex]log _{3} (x)+ log_{9} (x) =12[/tex]
[tex]log _{3} (x)+ log_{3^{2} } (x) =12[/tex]
we know that
[tex]log^{a} _{b} = \frac{loga}{logb}[/tex]
[tex]log _{3} (x)+ \frac{logx}{log3^{2} } =12[/tex]
Step(ii):-
Apply log xⁿ = nlogx
[tex]log _{3} (x)+\frac{1}{2} \frac{logx}{log3 } =12[/tex]
[tex]log _{3} (x)+ \frac{1}{2} log_{3 } (x) =12[/tex]
[tex]log _{3} (x)+ log_{3 } (x)^{\frac{1}{2} } =12[/tex] ( ∵ log xⁿ = nlogx)
Apply log(ab) = loga+logb
[tex]log _{3} (x (x^{\frac{1}{2} }) =12[/tex]
[tex]log _{3} ( (x^{\frac{3}{2} }) =12[/tex]
[tex]\frac{3}{2} log _{3} ( x) =12[/tex]
[tex]\frac{1}{2} log _{3} ( x) = 4[/tex]
[tex]log _{3} ( x) = 8[/tex]
we know that [tex]log _{b} ( x) = a[/tex] ⇒ x = bᵃ
∴ x = 3⁸
