Question

Help me on this question? log x-log 9=1

Answer 1

$\log { x } -\log { 9 } =1\\ \\ \log { \left( \frac { x }{ 9 } \right) } =1\\ \\ \log _{ 10 }{ \left( \frac { x }{ 9 } \right) } =1\\ \\ { 10 }^{ 1 }=\frac { x }{ 9 } \\ \\ 9\cdot 10=x\\ \\ x=90$

This is because:

$\log _{ a }{ \left( \frac { x }{ p } \right) } \\ \\ =\log _{ a }{ \left( \frac { { a }^{ m } }{ { a }^{ n } } \right) } \\ \\ =\log _{ a }{ \left( { a }^{ \left( m-n \right) } \right) } \\ \\ =\left( m-n \right) \cdot \log _{ a }{ a } \\ \\ =m-n\\ \\ =\log _{ a }{ x } -\log _{ a }{ p }$

Answer 2

   log(x) - log(9) = 1

Subtracting the logs of numbers gives you the log of
the quotient of the numbers.

     Log(x) - log(9)  is the log of  (x/9).

So the equation says:    log (x/9) = 1

Raise 10 to the power of each side:  10^(log of x/9) = 10^1

But  10^(log of x/9) is x/9, and 10^1 is 10.

So        x/9 = 10

             x  =  90