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 If  log3 [3^(x+1)  - 18] = 2 , find the value of x.

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Tushar Chandra eNotes educator | Certified Educator

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We are given that log(3) [3^(x+1)  - 18] = 2. We have to find x.

log(3) [3^(x+1)  - 18] = 2

=> [3^(x+1)  - 18] = 3^2

=> 3^x * 3 - 18 = 9

=> 3^x*3 = 27

=> 3^x = 27/3

=> 3^x = 9

=> 3^x = 3^2

As the base is the same, we can equate the exponent

=> x = 2

The required value is x = 2

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hala718 eNotes educator | Certified Educator

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Given the equation.

log3 [ 3^(x+1) -18) = 2

We need to solve for x.

We will use logarithm properties to solve for x.

First we will rewrite 18 = 3*6

log (3^x*3 - 3*6) = 2

Now factor 3.

log3 ( 3(3^x -6) = 2

Now we know that log a*b = log a + log b

==> log3 3 + log (3^x -6) = 2

==> 1 + log3 (3^x -6) = 2

==> log3 (3^x -6) = 1

rewrite into exponent form:

==> 3^x -6 = 3

==> 3^x = 9

==> 3^x = 3^2

==> x = 2

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