Question

Prove: $\sf{\log_3x+2·\log_x3=3}$​

Answer 1

Appropriate Question :-

Solve for x :-

$\rm :\longmapsto\: log_{3}(x) + 2 \: log_{x}(3) = 3$

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: log_{3}(x) + 2 \: log_{x}(3) = 3$

can be rewritten as

$\rm :\longmapsto\: log_{3}(x) + \dfrac{2}{log_{3}(x)} \: = 3$

$\: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because \: log_{x}(y) = \dfrac{1}{ log_{y}(x) } }}$

$\bf :\longmapsto\:Let \: log_{3}(x) \: =y - - - (1)$

So, we get

$\rm :\longmapsto\:y + \dfrac{2}{y} = 3$

$\rm :\longmapsto\: \dfrac{ {y}^{2} + 2}{y} = 3$

$\rm :\longmapsto\: {y}^{2} + 2 = 3y$

$\rm :\longmapsto\: {y}^{2} - 3y + 2 = 0$

$\rm :\longmapsto\: {y}^{2} - 2y - y + 2 = 0$

$\rm :\longmapsto\:y(y - 2) - 1(y - 2) = 0$

$\rm :\longmapsto\:(y - 2)(y - 1) = 0$

$\bf\implies \:y = 2 \: \: \: \: or \: \: \: \: y = 1$

Case :- 1

$\rm :\longmapsto\:When \: y \: = \: 2$

$\rm :\longmapsto\:log_{3}(x) = 2$

$\: \: \: \: \: \: \green{\boxed{ \bf \because \: log_{x}(y) = z \: then \: y = {x}^{z} }}$

$\rm :\longmapsto\:x = {3}^{2}$

$\bf\implies \:x = 9$

Case :- 2

$\rm :\longmapsto\:When \: y \: = \: 1$

$\rm :\longmapsto\:log_{3}(x) = 1$

$\rm :\longmapsto\:x = {3}^{1}$

$\: \: \: \: \: \: \green{\boxed{ \bf \because \: log_{x}(y) = z \: then \: y = {x}^{z} }}$

$\bf\implies \:x = 3$

Verification :-

When x = 9

Consider LHS

$\rm :\longmapsto\: log_{3}(9) + 2 \: log_{9}(3)$

$\rm : = \: \: \: log_{3}( {3}^{2} ) + 2 \: log_{ {3}^{2} }(3)$

$\rm : = \: \: \: 2 + 2 \times \dfrac{1}{2}$

$\rm = \: \: \: 2 + 1$

$\rm \: \: = \: \: 3$

Hence, LHS = RHS

Case :- 2

When x = 3

$\rm :\longmapsto\: log_{3}(3) + 2 \: log_{3}(3)$

$\: \rm \: = \: \: 1 + 2 \times 1$

$\: \rm \: = \: \: 1 + 2$

$\: \rm \: = \: \: 3$

Hence, LHS = RHS