$9^{3 c}+1=27^{3 c}-1

Rewrite the equation 9 3 c + 1 = 2 7 3 c − 1 as 3 6 c + 2 = 3 9 c .
Substitute x = 3 3 c to get x 3 − x 2 − 2 = 0 .
Factor the cubic equation as ( x − 2 ) ( x 2 + x + 1 ) = 0 , yielding the real root x = 2 .
Solve 2 = 3 3 c for c , resulting in c = 3 l n 3 l n 2 ​ .

Explanation

Problem Analysis We are given the equation 9 3 c + 1 = 2 7 3 c − 1 . Our goal is to solve for c . We will rewrite the equation, express both sides with a common base, and then solve the resulting equation.

Rewriting the Equation First, let's rewrite the equation as 9 3 c + 2 = 2 7 3 c . Since 9 = 3 2 and 27 = 3 3 , we can rewrite the equation in terms of base 3: ( 3 2 ) 3 c + 2 = ( 3 3 ) 3 c , which simplifies to 3 6 c + 2 = 3 9 c .

Substitution Now, let x = 3 3 c . Then x 2 = ( 3 3 c ) 2 = 3 6 c and x 3 = ( 3 3 c ) 3 = 3 9 c . Substituting these into the equation, we get x 2 + 2 = x 3 , which can be rearranged to x 3 − x 2 − 2 = 0 .

Solving the Cubic Equation We need to solve the cubic equation x 3 − x 2 − 2 = 0 . By observation, we can see that x = 2 is a root, since 2 3 − 2 2 − 2 = 8 − 4 − 2 = 2 . However, this is incorrect. Let's test x = 2 : 2 3 − 2 2 − 2 = 8 − 4 − 2 = 2 e q 0 . So x = 2 is not a root. Let's try to factor the cubic. If x = 2 is a root, then ( x − 2 ) must be a factor. Performing polynomial division, we have ( x 3 − x 2 − 2 ) / ( x − 2 ) = x 2 + x + 1 . Thus, x 3 − x 2 − 2 = ( x − 2 ) ( x 2 + x + 1 ) .

Finding the Roots Now we solve for the roots. First, we solve x − 2 = 0 , which gives x = 2 . Then, we solve x 2 + x + 1 = 0 using the quadratic formula: x = 2 − 1 ± 1 − 4 ​ ​ = 2 − 1 ± i 3 ​ ​ . Since x = 3 3 c , x must be a real and positive number. Therefore, we take the real root x = 2 .

Substituting Back Substitute x = 2 back into x = 3 3 c to get 2 = 3 3 c . Now, take the logarithm of both sides: ln 2 = ln ( 3 3 c ) = 3 c ln 3 .

Solving for c Finally, solve for c : c = 3 l n 3 l n 2 ​ . We can approximate this value using a calculator: c ≈ 3 ( 1.0986 ) 0.6931 ​ ≈ 3.2958 0.6931 ​ ≈ 0.2103 .

Final Answer Therefore, the solution for c is c = 3 l n 3 l n 2 ​ .


Examples
Exponential equations like this appear in various fields, such as calculating the growth of investments or modeling the decay of radioactive substances. For instance, if you invest an amount that grows exponentially, understanding how to solve such equations helps you determine the time it takes for your investment to reach a specific value. Similarly, in radioactive decay, these equations help in calculating the half-life of a substance, which is crucial in fields like medicine and environmental science. Let's say you have an investment that grows according to the formula A = P × ( 9 3 t ) , where A is the final amount, P is the principal, and t is the time. If you want to find out when the investment doubles ( A = 2 P ), you would solve 2 P = P \tims ( 9 3 t ) , which simplifies to 2 = 9 3 t . This is similar to the equation we solved, and the solution would give you the time t it takes for the investment to double.

Answered by GinnyAnswer | 2025-07-03

To solve the equation 9 3 c + 1 = 2 7 3 c − 1 , transform it to a cubic polynomial form and find the real root. The solution for c is derived as c = 3 l n 3 l n 2 ​ . Approximating gives c ≈ 0.2103 .
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Answered by Anonymous | 2025-07-04