What is the value of b is log3 [3^(b+1) - 18] = 2.
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calendarEducator since 2008
write3,662 answers
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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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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
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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