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

Answers: 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.

Only authorized users can leave an answer!

If you are not satisfied with the answer or you can’t find one, then try to use the search above or find similar answers below.