# (log3 x)^2+2log3x-24=0

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exordiumxAnswers: x = 1/729 or x = 81 ====== (log₃ x)² + 2log₃ x - 24 = 0 Notice how this seems to form some sort of quadratic equation. There is a tern with a variable that is squared log₃ x)², then a term with the same variable of the first degree (2log₃ x) and then a constant -24. This looks like ax² + bx + c = 0 Let u = log₃ x. Then this can be written as (log₃ x)² + 2log₃ x - 24 = 0 ⇒ (u)² + 2u - 24 = 0 We can solve this by factoring. Find two numbers that multiply to get -24 and add to get 2. These two numbers are 6 and -4. Therefore, the left-hand side factors into (u + 6)(u - 4) = 0 By the zero factor property, we find solutions by setting the factors to equal zero and then combining the solutions. So u + 6 = 0 or u - 4 = 0 Back-substituting u = log₃ x, we get log₃ x + 6 = 0 or log₃ x - 4 = 0 Isolating the logarithmic term log₃ x = -6 or log₃ x = 4 Use definition of the logarithm to convert these into exponential form. Since logₓ(a) = b ⇔ a = xᵇ, we convert so that these equations turn into x = 3⁻⁶ or x = 3⁴ x = 1/729 or x = 81 Check your solutions by putting them in the original equation. Doing that, both sides of the equation (log₃ x)² + 2log₃ x - 24 = 0 result in 0, indicating that these two numbers are the solutions to the given equation.