# Logarithmic Equation: `log_9 (2x^2 - 9) - log_9 8 = 1`

rakesh05 | High School Teacher | (Level 1) Assistant Educator

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Given logarithmic equation is `log_9(2x^2-9)-log_9 8=1` .

Now we know that `log_b m-log_b n=log_b (m/n)` . Also `1=log_b b` .

So,                       `log_9( (2x^2-9)/8)=log_9 9` .

Now we know that `log_b x=log_b yrArr x=y` .

So,                    `(2x^2-9)/8=9`

or,                   `2x^2-9=72`

or,                     `2x^2-81=0`

or,                       `x^2-81/2=0`

or,                  `(x^2-(9/sqrt2)^2)=0`

or,                 `(x-9/sqrt2)(x+9/sqrt2)=0` .

So,                  `x=+-9/sqrt2` .

This is the desired value of x.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The logarithmic equation `log_9 (2x^2 - 9) - log_9 8 = 1` has to be solved.

`log_9 (2x^2 - 9) - log_9 8 = 1`

=> `log_9(2x^2 - 9) - log_9 8 = 1 = log_9 9`

=> `log_9((2x^2 - 9)/8) = log_9 9`

`(2x^2 - 9)/8 = 9`

=> `2x^2 - 9 = 72`

=> `2x^2 = 81`

=> `x = +-sqrt(81/2)`

The solution of the equation is `x = +-sqrt(81/2)`