Explain the process the properties you have to use to solve the logarithmic equation: log3x + log34 - 2 log33 = 2
Accepted Solution
A:
First, we are going to isolate [tex]log3x[/tex] in the left side of the equation: [tex]log3x+log34-2log33=2[/tex] [tex]log3x=2+2log33-log34[/tex]
Next, we are going to use log of a power rule: [tex]nlogx=logx^{n}[/tex] [tex]log3x=2+2log33-log34[/tex] [tex]log3x=2+log33^{2}-log34[/tex] [tex]log3x=2+log1089-log34[/tex]
Next, we are going to use the log of a quotient rule: [tex]loba-logb=log \frac{a}{b} [/tex] [tex]log3x=2+log1089-log34 [/tex] [tex]log3x=2+log \frac{1089}{34} [/tex]
Next, we are going to use the rule: [tex]a=log_{b}b^{a}[/tex] [tex]log3x=2+log \frac{1089}{34} [/tex] [tex]log3x=log10^{(2+log \frac{1098}{34})[/tex] [tex]log3x=log[(10^{2})(10^{log \frac{1089}{34} })][/tex] [tex]log3x=log[(100)( \frac{1089}{34} )][/tex] [tex]log3x=log \frac{54450}{17} [/tex]
And last but not least, we are going to use the same base log rule: [tex]loga=logb [/tex] → [tex]a=b[/tex] [tex]log3x=log \frac{54450}{17} [/tex] [tex]3x= \frac{54450}{17} [/tex] [tex]x= \frac{54450}{(3)(17)} [/tex] [tex]x= \frac{18150}{17} [/tex]
We can conclude that the solution of our logarithmic equation is [tex]x= \frac{18150}{17} [/tex].