Find solution of: `log_8 (x+5) - log_8 (x-2) gt1` Express in set notation and interval notation 

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lemjay | High School Teacher | (Level 3) Senior Educator

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Notice that the logarithms at the left side has same base. So, express it as one logarithm using the rule `log_bM-log_bN=log_b(M/N).`  .


Then, convert this equation to exponential form.



Then, subtract both sides by 8 to have a zero at the right side of the equation.


Then, express them with same denominator to simplify the left side.





Now, that the equation is in simplified form, determine the critical numbers by setting the numerator and denominator equal to zero.

`-7x + 21 = 0`     and     `x - 2 =0`
`x=3`                                    `x=2`

Since there are only two critical numbers, then there are three possible intervals as solution to the given equation. These are `xlt2`  ,  `2ltxlt3`  and  `xgt3` .

To determine which of them is a solution to the given equation, assign a test value for each interval. Then, plug-in them to the original equation. The interval which satisfy the inequality equation is the solution.

For the first first interval x < 2, let x=0.



`log_8 5-log_8(-2)gt1 `
Take note that in logarithm, negative argument is not allowed.Since the argument of the second logarithm is negative, then the expression is invalid. So, x<2 is not a solution.

For the second interval 2<x<3, let x=2.5.


`log_8 7.5-log_8 0.5gt1`

`1.3gt1` (True)

Hence, the interval 2<x<3 is a solution to the given equation.

And for the third interval x > 3, let x=4.


`log_8 9 - log_8 2 gt 1`

`0.7gt1` (False)

So, the third interval is not a solution.

Therefore, the solution to the equation `log_8(x+5)-log_8(x-2)gt1` is `2ltxlt3` . Its interval notation is `(2,3)` .

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oldnick | (Level 1) Valedictorian

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From the one to one values of logatirm funtion:






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