If log3 (12n-1) - log3 (2n+7) = 2 find the value of n.

Expert Answers
hala718 | Certified Educator

log3 (12n-1) - log3 (2x+7) = 2

First we will find the domain.

==> 12n-1 > 0 ==> n> 1/12

==> 2n+7 > 0 ==> n > -7/2

Then The domain is n > 1/12 .............(1)

Now we will solve.

We know that log a - log b= log a/b.

==> log3 (12n-1)/(2n+7) = 2

Now we will rewrite using exponent form.

==> (12n-1)/(2n+7) = 3^2

==> (12n-1)/(2n+7) = 9

Now we will multiply by (2n+7)

==> 12n-1 = 9(2n+7)

==> 12n -1 = 18n + 63

==> -6n = 64

==> n= -64/6 = -10.6

But -10.6 does not belong to the domain.

Then the equation has no real solution.

justaguide | Certified Educator

It is given that log3 (12n-1) - log3 (2n+7) = 2

Use the property of logarithms: log a - log b = log(a/b)

log3 (12n-1) - log3 (2n+7) = 2

=> log3[(12n-1)/(2n+7)] = 2

=> (12n-1)/(2n+7)] = 3^2

=> (12n-1)/(2n+7)] = 9

=> 12n - 1 = 9(2n + 7)

=> 12n - 1 = 18n + 63

=> 6n = -64

=> n = -64/6

But log (12n - 1) is not defined for a negative value of n.

The equation does not have any solution for n

tonys538 | Student

The equation `log_3 (12n-1) - log_3 (2n+7) = 2` has to be solved for n.

Use the relation, `log a - log b = log(a/b)` .

`log_3 (12n-1) - log_3 (2n+7) = 2` gives:

`log_3((12n-1)/(2n+7)) = 2`

Now `log_a b = c` implies `b = a^c` .

`log_3((12n-1)/(2n+7)) = 2` implies `(12n-1)/(2n+7)) = 3^2 = 9`

`12n - 1 = 9*(2n+7)`

`12n - 1 = 18n + 63`

`6n = -64`

`n = -64/6`

`n = -32/3`

But `log_3 (12n-1)` and `log_3 (2n+7)` are not defined for `n = -32/3` . The given equation has no solution.

Access hundreds of thousands of answers with a free trial.

Ask a Question
Popular Questions