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QuestionSolve 32n+1=2(21−3n)3^{2 n+1}=2\left(2^{1-3 n}\right) for 'n'.

Studdy Solution

STEP 1

Assumptions1. The given equation is 3n+1=(1−3n)3^{ n+1}=\left(^{1-3 n}\right). We are solving for the variable 'n'

STEP 2

First, we can simplify the right side of the equation.
2(21−n)=21+1−n=22−n2\left(2^{1- n}\right) =2^{1+1-n} =2^{2-n}

STEP 3

Now, we rewrite the equation with the simplified right side.
32n+1=22−3n3^{2 n+1}=2^{2-3n}

STEP 4

Next, we can take the logarithm base2 of both sides of the equation. This is because the logarithm is the inverse of exponentiation, and it will allow us to bring down the exponents.
log⁡2(32n+1)=log⁡2(22−3n)\log2\left(3^{2 n+1}\right) = \log2\left(2^{2-3n}\right)

STEP 5

Now we can use the property of logarithms that allows us to bring the exponent down as a coefficient.
(2n+1)log⁡2(3)=2−3n(2 n+1)\log2(3) =2-3n

STEP 6

Next, we can distribute the log⁡2(3)\log2(3) on the left side of the equation.
2nlog⁡2(3)+log⁡2(3)=2−3n2n\log2(3) + \log2(3) =2-3n

STEP 7

Now, we can rearrange the equation to isolate the terms with 'n' on one side and the constants on the other side.
2nlog⁡2(3)+3n=2−log⁡2(3)2n\log2(3) +3n =2 - \log2(3)

STEP 8

We can factor out 'n' on the left side of the equation.
n(2log⁡2(3)+3)=2−log⁡2(3)n(2\log2(3) +3) =2 - \log2(3)

STEP 9

Finally, we can solve for 'n' by dividing both sides of the equation by the coefficient of 'n'.
n=2−log⁡2(3)2log⁡2(3)+3n = \frac{2 - \log2(3)}{2\log2(3) +3}This is the solution for 'n'.

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