Example: Solve following equations.
2 log_{9} (√x) - log_{9} (6x -1) = 0
Solution
Along with this equation there are two logarithms only in the equation thus it's easy to get on one either side of the equal sign. We will also have to deal with the coefficient in front of the first term.
log9 (√x)^{2} = log_{9} (6x -1)
log_{9 }x = log_{9 }(6x -1)
Now that we've got two logarithms along with the similar base & coefficients of 1 on either side of the equal sign we can drop the logs & solve.
x = 6 x -1
1 = 5x ⇒ x = 1/5
Now, we do have to worry if this solution will generate any negative numbers or zeroes in the logarithms hence the next step is to plug this into the original equation & see if it does.
2 log_{9} (√(1/5) - log_{9}( 6 ( 1 /5) -1) = 2 log_{9} (√1/5 ) - log9 ( √(1 /5)) = 0
Note that we don't have to go all the way out with the check here. We just have to ensure that once we plug in the x we don't contain any negative numbers or zeroes in the logarithms. Since we don't in this case we have the solution, it is x=(1/5)