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Solve for X in the equation X^3 - X^0 = 26
Published on November 13, 2024
Algebra
The solution to the equation X^3 - X^0 = 26 is X = 3.
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Published on November 13, 2024
The solution to the equation X^3 - X^0 = 26 is X = 3.
The equation (x-1)^(|x|-4) = 1 has solutions x = -4, 0, 2, and 4.
The equation x^2 - 2x = 0 has two solutions, x = 0 and x = 2, not just x = 2.
The solution for b in the equation 2^(b^2+2) = 64 is b = 2.
The multiplication of the two given matrices results in a 2x3 matrix with entries calculated by multiplying corresponding rows and columns of the matrices and summing the products.
The equation x^x + 6/x = 7 has two solutions, x = 1 and x = 2, as demonstrated by direct substitution.
The solution to the equation 3^(x+2) = 2^(x+3) is found by taking the natural logarithm of both sides, applying logarithm properties, isolating x, and simplifying to an expression involving natural logarithms of 2 and 3. The approximate value of x is then calculated.
The expression $((i+1)(i-1))^2$ simplifies to 4.
The equation 3^(x^2)/9^x = 81 is solved by rewriting the equation with the same base, simplifying the denominator, applying the quotient rule for exponents, equating the exponents, rearranging into a quadratic equation, and finally using the quadratic formula to find the two solutions, x = 1 + √5 and x = 1 - √5.
The equation x^2 + 3x + 7 = 6/(x^2 + 3x + 2) is solved by substituting u = x^2 + 3x, resulting in a quadratic equation in u. Factoring the quadratic equation yields two possible values for u, which are then substituted back into the substitution to find the solutions for x using the quadratic formula. One case results in two real solutions, while the other yields no real solutions.
The solution to the equation \sqrt{7 + \frac{3}{\sqrt{x}}} = 7 - \frac{9}{x} is x = \frac{9}{4}.