Elements of Algebra

270 ELEMENTS SECT. IV. 3. Given $y^3 + 24y = 250$, to find the value of $y$. Ans. $y = 5.05$. 4. Given $y^6 - 3y^4 - 2y^2 - 8 = 0$, to find $y$. Ans. $y = 2$. Log. $x = \frac{1}{2}$ log. $\frac{4p}{3} +$ log. cos. $\frac{z}{3} - 10$, Log. $x = \frac{1}{2}$ log. $\frac{4p}{3} +$ log. cos. $(60^\circ - \frac{z}{3}) - 10$, Log. $x = \frac{1}{2}$ log. $\frac{4p}{3} +$ log. cos. $(60^\circ + \frac{z}{3}) - 10$, Taking the value of $x$, answering to log. $x$, positively in the first equation, and negatively in the two latter. 4. $x^3 - px + q = 0$. This form, like the former, has also two cases, according as $\frac{2}{q}(\frac{p}{3})^{\frac{3}{2}}$ is less, or greater than 1. In the 1st case, put $\frac{2}{q}(\frac{p}{3})^{\frac{3}{2}} = \cos. z$, And $\sqrt[3]{(\tan.(45^\circ - \frac{1}{2}z))} = \tan. u$, as before; Then $x = -2\sqrt{\frac{p}{3}}$ cosec. $2u$. Or, putting $10 + \frac{1}{2}$ log. $\frac{p}{3} - \log. \frac{q}{2} = \log. \cos. z$, and $\frac{1}{3}\{\log.(\tan. 45^\circ - \frac{1}{2}z) + 20\} = \log. \tan. u$; Then, $-\log. x = 10 + \log. \frac{4p}{3} - \log. \sin. 2u$. In the 2d case, put $\frac{q}{2}(\frac{3}{p})^{\frac{3}{2}} = \cos. z$, and $x$ will have the 3 following values: $x = -2\sqrt{\frac{p}{3}} \times \cos. \frac{z}{3}$ $x = +2\sqrt{\frac{p}{3}} \times \cos. (60^\circ - \frac{z}{3})$ $x = +2\sqrt{\frac{p}{3}} \times \cos. (60^\circ + \frac{z}{3})$. Or, Log. $x = \frac{1}{2}$ log. $\frac{4p}{3} +$ log. cos. $\frac{z}{3} - 10$, Log. $x = \frac{1}{2}$ log. $\frac{4p}{3} +$ log. cos. $(60^\circ - \frac{z}{3}) - 10$, Log. $x = \frac{1}{2}$ log. $\frac{4p}{3} +$ log. cos. $(60^\circ + \frac{z}{3}) - 10$,