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Question Number 160879 by amin96 last updated on 08/Dec/21

Commented by cortano last updated on 08/Dec/21

$$\iff 27.{\mathrm{x}}^{\log {}_3\left({\mathrm{x}}^{\frac{1}{3}}\right)}={\mathrm{x}}^{\frac{10}{3}}$$$$\iff \log {}_3\left(27\right)+\frac{1}{3}{\left(\log {}_3\left(\mathrm{x}\right)\right)}^2=\frac{10}{3}\log {}_3\left(\mathrm{x}\right)$$$$\iff 3+\frac{1}{3}{\mathrm{t}}^2-\frac{10}{3}\mathrm{t}=0;[\mathrm{t}=\log {}_3\left(\mathrm{x}\right)]$$$$\iff {\mathrm{t}}^2-10\mathrm{t}+9=0$$$$\iff (\mathrm{t}-1)(\mathrm{t}-9)=0$$$$\to \{\begin{array}{l}\mathrm{t}=1\Rightarrow \mathrm{x}=3^1=3\\ \mathrm{t}=9\Rightarrow \mathrm{x}=3^9\end{array}$$

Answered by MathsFan last updated on 08/Dec/21

$$\mathbf{let}{\mathbf{log}}_{27}\mathbf{x}=\mathbf{u}\to \mathbf{x}={27}^{\mathbf{u}}$$$$27\bullet {\left({27}^{\mathbf{u}}\right)}^{\mathbf{u}}=\sqrt[3]{{\left({27}^{\mathbf{u}}\right)}^{10}}$$$$27\bullet {27}^{{\mathbf{u}}^2}={27}^{10\mathbf{u}/3}$$$${\mathbf{u}}^2+1=\frac{10\mathbf{u}}{3}$$$$3{\mathbf{u}}^2-10\mathbf{u}+3=0$$$${\mathbf{u}}_1=3\mathbf{and}{\mathbf{u}}_2=\frac{1}{3}$$$${\mathbf{x}}_1={27}^3=19683$$$${\mathbf{x}}_2=3$$

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