Question-50802

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 \Leftrightarrow (x^{2} - x - 6) l n (4) = l n (7) \Leftrightarrow

x^{2} - x - 6 = \frac{l n (7)}{l n) 4)} \Leftrightarrow x^{2} - x - 6 = {l o g}_{4} (7) \Rightarrow

x^{2} - x - 6 - {l o g}_{4} (7) = 0

Δ = 1 - 4 (- 6 - {l o g}_{4} (7)) = 25 + 4 {l o g}_{4} (7) > 0 \Rightarrow

x_{1} = \frac{1 + \sqrt{25 + 4 {l o g}_{4} (7)}}{2} a n d

x_{2} = \frac{1 - \sqrt{25 + 4 {l o g}_{4} (7)}}{2}

Answered by hassentimol last updated on 20/Dec/18

Answers are :

S = {x = 0.5 \underline{+} \frac{\sqrt{25 + 4 \times \log_{4} (7)}}{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} (7) = x^{2} - x - 6

x^{2} - x - (6 + \log_{4} (7)) = 0

to find x_{1, 2} : \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

x_{1, 2} = \frac{1 \pm \sqrt{1^{} + 4 {(6 + \log_{4} (7))}^{}}}{2}

x_{1, 2} = \frac{1 \pm \sqrt{25 + 4 \log_{4} {(7)}^{}}}{2}

Facebook Tweet Pin