Given:
The equation is
[tex]\log_318+\log_33-\log_3x=3[/tex]
To find:
The solution for the given equation.
Solution:
We have,
[tex]\log_318+\log_33-\log_3x=3[/tex]
Using the properties of logarithm, we get
[tex]\log_3(18\times 3)-\log_3x=3[/tex] [tex]\left[\because \log(ab)=\log a+\log b\right][/tex]
[tex]\log_3(54)-\log_3x=3[/tex]
[tex]\log_3\left(\dfrac{54}{x}\right)=3[/tex] [tex]\left[\because \log(\dfrac{a}{b})=\log a-\log b\right][/tex]
[tex]\dfrac{54}{x}=3^3[/tex] [tex][\because \log_ab=x\Leftrightarrow b=a^x][/tex]
On further simplification, we get
[tex]\dfrac{54}{x}=27[/tex]
[tex]\dfrac{54}{27}=x[/tex]
[tex]2=x[/tex]
Therefore, the value of x is 2.
