If log3 [3^(x+1) - 18] = 2 , find the value of x.
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calendarEducator since 2010
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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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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
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