Question

pls someone help to do this one​

Answer 1

Step-by-step explanation:

${ \sqrt{3} }^{x - 3} = { \sqrt[4]{3} }^{x + 1} \\ \ { \sqrt[4]{3} }^{x + 1} can \: be \: writen \: as \: { \sqrt{3} }^{ \frac{x + 1}{2} } \\ { \sqrt{3} }^{x - 3} = { \sqrt[4]{3} }^{x + 1} \\ \ { \sqrt[4]{3} }^{x + 1} can \: be \: writen \: as \: { \sqrt{3} }^{ \frac{x + 1}{2} } \\ so \: \\ { \sqrt{3} }^{x - 3} = { \sqrt{3} }^{ \frac{x + 1}{2} } \\ x - 3 = \frac{x + 1}{2} \\ 2(x - 3) = x + 1 \\ 2x - 6 = x + 1 \\ 2x - x = 1 + 6 \\ x = 7 \\ so \: x \: is \: 7\\</p><p>can\:you\:please\:write\:2nd\:que. \:properly$