Question

q If x² + y² =25xy, then prove that a
that 2 log(xty)
Blog 3 et log2 tlogy​

Answer 1

Answer:

${ \underline{ \large{ \pmb{ \sf{Correct \: Question... }}}}}$

• If x² + y² = 25xy , Then prove that 2log(x + y) = 3log3 + logx + logy

${ \underline{ \large{ \pmb{ \sf{Solution... }}}}}$

$: { \implies{ \sf{ {x}^{2} + {y}^{2} = 25xy }}} \\ \\ { \sf{Adding \: 2xy \: on \: both \: sides}} \\ \\ : { \implies{ \sf{ {x}^{2} + {y}^{2} + 2xy = 25xy + 2xy}}} \\ \\ : { \implies{ \sf{ {(x + y)}^{2} = 27xy}}} \\ \\ { \sf{Taking \: Log \: on \: both \: sides}} \\ \\ : { \implies{ \sf{log {(x + y)}^{2} = log27xy }}} \\ \\ : { \implies{ \sf{2log(x + y) = log27 + logx + logy}}} \\ \\ : { \implies{ \sf{2log(x + y) = log {3}^{3} + logx + logy}}} \\ \\ : { \implies{ \sf{2log(x + y) =3 log 3 + logx + logy}}} \\ \\ \therefore { \pmb{ \sf{Hence \: Proved}}}$

More to know :

• ${ \sf{log {a}^{n} = nloga }}$

• ${ \sf{log(ab) = loga + logb}}$

• ${ \sf{log1 = 0}}$

• ${ \sf{ log_{a}(a) = 1}}$

• ${ \sf{log \frac{a}{b} = loga - logb }}$

Answer 2

Answer:

CorrectQuestion...

CorrectQuestion...

If x² + y² = 25xy , Then prove that 2log(x + y) = 3log3 + logx + logy

{ \underline{ \large{ \pmb{ \sf{Solution... }}}}}

Solution...

Solution...

\begin{gathered} : { \implies{ \sf{ {x}^{2} + {y}^{2} = 25xy }}} \\ \\ { \sf{Adding \: 2xy \: on \: both \: sides}} \\ \\ : { \implies{ \sf{ {x}^{2} + {y}^{2} + 2xy = 25xy + 2xy}}} \\ \\ : { \implies{ \sf{ {(x + y)}^{2} = 27xy}}} \\ \\ { \sf{Taking \: Log \: on \: both \: sides}} \\ \\ : { \implies{ \sf{log {(x + y)}^{2} = log27xy }}} \\ \\ : { \implies{ \sf{2log(x + y) = log27 + logx + logy}}} \\ \\ : { \implies{ \sf{2log(x + y) = log {3}^{3} + logx + logy}}} \\ \\ : { \implies{ \sf{2log(x + y) =3 log 3 + logx + logy}}} \\ \\ \therefore { \pmb{ \sf{Hence \: Proved}}}\end{gathered}

:⟹x

2

+y

2

=25xy

Adding2xyonbothsides

:⟹x

2

+y

2

+2xy=25xy+2xy

:⟹(x+y)

2

=27xy

TakingLogonbothsides

:⟹log(x+y)

2

=log27xy

:⟹2log(x+y)=log27+logx+logy

:⟹2log(x+y)=log3

3

+logx+logy

:⟹2log(x+y)=3log3+logx+logy

∴

HenceProved

HenceProved

More to know :

{ \sf{log {a}^{n} = nloga }}loga

n

=nloga

{ \sf{log(ab) = loga + logb}}log(ab)=loga+logb

{ \sf{log1 = 0}}log1=0

{ \sf{ log_{a}(a) = 1}}log

a

(a)=1

{ \sf{log \frac{a}{b} = loga - logb }}log

b

a

=loga−logb