Question

Express $x^{\frac{3}{4}}({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4}=\sqrt{3}$​ in terms of t where t = ${\log }_3$ x.

• A. + 7t + 2 = 0
• B. + + 9t + 2 = 0
• C. + - 5t - 2 = 0
• D. + - 5t - 2 = 0

Answer 1

The correct option is C

+ - 5t - 2 = 0

Given equation is $x^{\frac{3}{4}}({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4}=\sqrt{3}$

Taking log on both sides at base 3

Taking log on both sides of equation 1 at base 3

${\log }_3\left\{x^{\frac{3}{4}({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4}}\right\}={\log }_3{3}^{\frac{1}{2}}$

$\left(\frac{3}{4}\right({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4}){\log }_{3}x=\frac{1}{2}$

Substitute ${\log }_{3}x$ = t

$\frac{3}{4}{t}^3$ + $t^2$ - $\frac{5}{4}t$ = $\frac{1}{2}$

$3t^2$ + $4t^2$ - 5t = 2

$3t^3$ + $4t^2$ - 5t - 2 = 0