If 20 (3−2x2)=(405) (3x2−2), then x=

Given 20 (3−2x2)= (405) 3x2−2⇒ 20 3−2x2=(40×40×5) 3x2−2⇒ 203−2x2= (20)32 (3x2−2)∴ 3−2x2=32(3x2−2)⇒ 6−4x2=9x2−6⇒ 13x2=12⇒ x2= 1213⇒ x= ± 1213

If 20 (3−2x2)=(405) (3x2−2), then x=

Given 20 (3−2x2)= (405) 3x2−2⇒ 20 3−2x2=(40×40×5) 3x2−2⇒ 203−2x2= (20)32 (3x2−2)∴ 3−2x2=32(3x2−2)⇒ 6−4x2=9x2−6⇒ 13x2=12⇒ x2= 1213⇒ x= ± 1213

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$I f 20^{(3 - 2 x^{2})} = (40 \sqrt{5})^{(3 x^{2} - 2)}, t h e n x =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\pm \sqrt{\frac{13}{12}}$

$\pm \sqrt{\frac{12}{13}}$

$\pm \sqrt{\frac{4}{5}}$

$\pm \sqrt{\frac{5}{4}}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} G i v e n 20^{(3 - 2 x^{2})} = {(40 \sqrt{5})}^{3 x^{2} - 2} \\ \Rightarrow 20^{3 - 2 x^{2}} = (\sqrt{40 \times 40 \times 5})^{3 x^{2} - 2} \\ \Rightarrow 20^{3 - 2 x^{2}} = {(20)}^{\frac{3}{2} (3 x^{2} - 2)} \\ ∴ 3 - 2 x^{2} = \frac{3}{2} (3 x^{2} - 2) \\ \Rightarrow 6 - 4 x^{2} = 9 x^{2} - 6 \\ \Rightarrow 13 x^{2} = 12 \\ \Rightarrow x^{2} = \frac{12}{13} \\ \Rightarrow x = \pm \sqrt{\frac{12}{13}} \end{array}$