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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`log_8(x+5)-log_8(x-2)gt1`

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).`  .

`log_8((x+5)/(x-2)gt1`

Then, convert this equation to exponential form.

`(x+5)/(x-2)gt8^1`

`(x+5)/(x-2)gt8`

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

`(x+5)/(x-2)-8gt0`

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

`(x+5)/(x-2)-(8(x-2))/(x-2)gt0`

`((x+5)-8(x-2))/(x-2)gt0`

`(x+5-8x+16)/(x-2)gt0`

`(-7x+21)/(x-2)gt0`

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(x+5)-log_8(x-2)gt1`

`log_8(0+5)-log_(0-2)gt1`

`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(2.5+5)-log_8(2.5-2)gt1`

`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(4+5)-log_8(4-2)gt1`

`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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`log_8(x+5)-log_8(x-2)>1`

`log_8(x+5)>1+log_8(x-2)`

`log_8(x+5)>log_8(8)+log_8(x-2)`

`log_8(x+5)>log[8(x-2)]`

From the one to one values of logatirm funtion:

`x+5>8(x-2)`

`x+5>8x-16`

`7x-21<0`

`7x<21`

`x<3`

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