Question

☺Try these 2 questions guys ☺

↔ Logarithm ↔

Circled Questions only ☺

Answer 1

HELLO DEAR,

WE KNOW THAT:-

BASIC FORMULA:-
IF equation is like this
a^x = Y

We can write like that , log_{a} y = x


GIѴƐɳ TɦⱭT :-

$log _{3}( {3}^{x} - 8) = 2 - x \\ \\ \\ {3}^{2 - x} = {3}^{x} - 8 \\ \\ \\ {3}^{x} - {3}^{2 - x} = ( 9 - 1) \\ \\ \\ {3}^{x} - {3}^{2 - x} = {3}^{2} - {3}^{0}$

Ѳɳ CѲⱮPⱭƦIɳG ƁѲTɦ รIƊƐ

WƐ GƐT,

X = 2



ѕє¢ση∂ qυєѕтιση,


$log_{2} \frac{(9 - {2}^{x}) }{3 - x} = 1 \\ \\ \\ log_{2} (9 - {2}^{x} ) = (3 - x) \\ \\ \\ 9 - {2}^{x} = {2}^{3 - x} \\ \\ \\ {2}^{3 - x} + {2}^{x} = (8 + 1) \\ \\ \\ {2}^{3 - x} + {2}^{x} = {2}^{3} + {2}^{0} \\ \\ \\ we \: get \: \\ \\ 3 - x = 3 \: \: or \: \: x \: = 0 \\ \\ \\ x = > 3 - 3 = 0 \: \: \: or \: \: \: x = 0$





I HOPE ITS HELP YOU DEAR,
THANKS