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Question Number 50802 by Pk1167156@gmail.com last updated on 20/Dec/18

Commented by Pk1167156@gmail.com last updated on 20/Dec/18

help soon please!

Commented by Abdo msup. last updated on 20/Dec/18

$$4^{x^2-x-6}=7\iff (x^2-x-6)ln\left(4\right)=ln\left(7\right)\iff$$$$x^2-x-6=\frac{ln\left(7\right)}{ln)4)}\iff x^2-x-6={log}_4\left(7\right)\Rightarrow$$$$x^2-x-6-{log}_4\left(7\right)=0$$$$\mathrm{\Delta }=1-4(-6-{log}_4\left(7\right))=25+4{log}_4\left(7\right)>0\Rightarrow$$$$x_1=\frac{1+\sqrt{25+4{log}_4\left(7\right)}}{2}and$$$$x_2=\frac{1-\sqrt{25+4{log}_4\left(7\right)}}{2}$$

Answered by hassentimol last updated on 20/Dec/18

$$\mathrm{Answers}\mathrm{are}:$$$$\mathbb{S}=\{x=0.5\underset{-}{+}\frac{\sqrt{25+4\times {\log }_4\left(7\right)}}{2}\}$$

Commented by Pk1167156@gmail.com last updated on 20/Dec/18

solution please!

Answered by afachri last updated on 20/Dec/18

$$4^{x^2-x-6}=7$$$$\log {}_4\left(7\right)=x^2-x-6$$$$x^2-x-(6+\log {}_4\left(7\right))=0$$$$\mathbf{to}\mathbf{find}{\mathit{x}}_{1,2}:\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$$$x_{1,2}=\frac{1\pm \sqrt{1^{}+4{(6+{\log }_4\left(7\right))}^{}}}{2}$$$$x_{1,2}=\frac{1\pm \sqrt{25+4\log {}_4{\left(7\right)}^{}}}{2}$$

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