Question

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Answer 1

Given:-

• $\sf2 log_{1 0 }(x) + \dfrac{1}{2} log_{10}(y) = 1$

To express:-

• y in terms of x

Answer:-

Given that,

$\sf2 log_{1 0 }(x) + \dfrac{1}{2} log_{10}(y) = 1$

${\red{\bigstar}} \boxed{\sf{alog_b(c) = log_b(c)^a}}$

$\sf \: \longrightarrow log_{10}( {x}^{2} ) + log_{10}( {y}^{ 1/2} ) = 1$

$\sf \: \longrightarrow log_{10}( {x}^{2} ) + log_{10}( \sqrt{y} ) = 1$

${\blue{\bigstar}} \boxed{\sf{log_b(a) + log_b(c) = log_b(ac)}}$

$\sf \longrightarrow log_{10}( {x}^{2} \sqrt{y} ) = 1$

Writing this in exponential form,

$\sf \longrightarrow {x}^{2} \sqrt{y} = 10$

$\sf \longrightarrow \sqrt{y} = \dfrac{10}{ {x}^{2} }$

Squaring both sides

$\sf \longrightarrow \underline{ \underline{ \sf{ \green{y = \dfrac{100}{ {x}^{4} } }}}}$

Answer 2

Answer:

Given that,

\sf2 log_{1 0 }(x) + \dfrac{1}{2} log_{10}(y) = 12log

10

(x)+

2

1

log

10

(y)=1

{\red{\bigstar}} \boxed{\sf{alog_b(c) = log_b(c)^a}}★

alog

b

(c)=log

b

(c)

a

\sf \: \longrightarrow log_{10}( {x}^{2} ) + log_{10}( {y}^{ 1/2} ) = 1⟶log

10

(x

2

)+log

10

(y

1/2

)=1

\sf \: \longrightarrow log_{10}( {x}^{2} ) + log_{10}( \sqrt{y} ) = 1⟶log

10

(x

2

)+log

10

(

y

)=1

{\blue{\bigstar}} \boxed{\sf{log_b(a) + log_b(c) = log_b(ac)}}★

log

b

(a)+log

b

(c)=log

b

(ac)

\sf \longrightarrow log_{10}( {x}^{2} \sqrt{y} ) = 1⟶log

10

(x

2

y

)=1

Writing this in exponential form,

\sf \longrightarrow {x}^{2} \sqrt{y} = 10⟶x

2

y

=10

\sf \longrightarrow \sqrt{y} = \dfrac{10}{ {x}^{2} }⟶

y

=

x

2

10

Squaring both sides

\sf \longrightarrow \underline{ \underline{ \sf{ \green{y = \dfrac{100}{ {x}^{4} } }}}}⟶

y=

x

4

100