What is the value of b is log3 [3^(b+1) - 18] = 2.

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hala718 eNotes educator | Certified Educator

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Given the equation:

log3 [3^(b+1) - 18] = 2.

We need to find the value of b.

First we will rewrite into the exponent form.

==> (3^(b+1) - 18) = 9

==> Now we will add 18 to both sides.

==> 3^(b+1) = 9 +18= 27

Now we will rewrite 27 = 3^3

==> 3^(b+1) = 3^3

Now that the bases are equal, then the powers are equal too.

==> b+ 1 = 3

==> b= 2

 

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justaguide eNotes educator | Certified Educator

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We have log(3) [3^(b+1) - 18] = 2

log(3) [3^(b+1) - 18] = 2

taking the anti-log of both the sides

3^(b+1) - 18 = 3^2

=> 3^( b + 1)  - 18 = 9

=> 3^( b+ 1) = 27

=> 3^( b + 1) = 3^3

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

b + 1 = 3

=> b = 2

The required value of b is 2.

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