Question

If ${\log }_{y}x+{\log }_{x}y=7$ then $({\log }_{y}x{)}^2+({\log }_{x}y{)}^2$ is equal to

• A. $43$
• B. $45$
• C. $47$
• D. $49$

Answer 1

The correct option is C $47$

$lo{g}_{y}x+lo{g}_{x}y=7$ then $(lo{g}_{y}x{)}^2+(lo{g}_{x}y{)}^2$

$\to lo{g}_{y}x+lo{g}_{x}y=7$

square both side

$(lo{g}_{y}x+lo{g}_{x}y{)}^2=(7{)}^2$

$(lo{g}_{y}x{)}^2+(lo{g}_{x}y{)}^2+2.lo{g}_{y}x.lo{g}_{x}y=49$

$(lo{g}_{y}x{)}^2+(lo{g}_{x}y{)}^2+2.lo{g}_{y}x.\frac{1}{lo{g}_{y}x}=49$

$(lo{g}_{y}x{)}^2+(lo{g}_{x}y{)}^2+2=49$

$(lo{g}_{y}x{)}^2+(lo{g}_{x}y{)}^2=47$