Question

if 20^((3-2x^(2)))=(40 sqrt(5))^((3x^(2)-2)) then x=​

Answer 1

Given:

20^ ( (3 - 2x^ (2) ) ) = (40 sqrt(5) )^ ( (3x^(2) - 2) )

To find:

if 20^ ( (3 - 2x^ (2) ) ) = (40 sqrt(5) )^ ( (3x^(2) - 2) ) then x=​

Solution:

From given, we have,

20^ ( (3 - 2x^ (2) ) ) = (40 sqrt(5) )^ ( (3x^(2) - 2) )

$20^{\left(\left(3-2x^{\left(2\right)}\right)\right)}=\left(40\sqrt{5}\right)^{\left(\left(3x^{\left(2\right)}-2\right)\right)}$

but $\left(40\sqrt{5}\right)^{3x^2-2}=\left(20^{\frac{3}{2}}\right)^{3x^2-2}$

$20^{3-2x^2}=20^{\frac{3}{2}\left(3x^2-2\right)}$

as bases are same, equating the exponential terms, we get,

$3-2x^2=\dfrac{3}{2}\left(3x^2-2\right)$

$3-2x^2=\dfrac{9}{2}x^2-3$

$-\dfrac{13}{2}x^2=-6$

-13x^2=-12

$x^2=\dfrac{12}{13}$

Therefore, we get the solution

$x=\dfrac{2\sqrt{39}}{13},\:x=-\dfrac{2\sqrt{39}}{13}$