Question

Let (xo, yo be the solution of the following
equations
(2x)2 =(3)3
3km = 2 km
Then xo is​

Answer 1

QUESTION :-

$\blue{ \sf Let \: (x_0,y_0) \: be \: the \: solution \: of \: the \: following}$

$\blue{ \sf equations :}$

$\: \: \: \: \: \: \: \: \: \: \: \blue{ \sf \: ( { {2x)}^{1n (2 )} } = {(3y)}^{ln(3)} } \\ \: \: \: \: \: \: \: \: \: \: \blue{ \sf {3}^{ln(x)} = {2}^{ln(y)} }$

${ \blue{ \sf{Then \: x_0 \: is:}}}$

EXPLANATION :-

${ \longrightarrow{ \blue{ \sf{( {2x)}^{ln(2)} = ( {3y)}^{ln(3)} }}}}$

Take ln(log) on both sides,

${ \longrightarrow{ \sf{ \: ln( {2x)}^{ln(2)} = ln( {3y)}^{ln(3)} }}}$

According to the property of log, we can shift the value of power to its front.

${ \longrightarrow{ \sf{ \: (ln \: 2)(ln \: 2x) = (ln \: 3)(ln \: 3y)}}}$

${ \longrightarrow{ \sf{ \: (ln \: 2) \: [ln \: 2 + ln \: x] = (ln \: 3) \: [ln \: 3 + ln \: y]}}}$

Open the brackets, and multiply with ln2 and ln3.

${ \longrightarrow{ \sf{ \: ( {ln \: 2)}^{ 2} + (ln2)(lnx) = ( {ln \: 3)}^{ 2} + (ln3)(lny) }}}$

${ \longrightarrow{ \sf{ \: ( {ln \: 2)}^{2} - ( {ln \: 3)}^{2} = (ln3)(lny) - ( ln2)(lnx) \: \: \: \: \: \: \: \: \: { \red{..(i)}}}}}$

• solve another equation

${ \longrightarrow{ \blue{\sf{ \: {3}^{ln x} = {2}^{lny} }}}}$

Take ln(log) on both sides,

${ \longrightarrow{ \sf{ \:ln( {3)}^{lnx} = ln( {2)}^{lny} }}}$

${ \longrightarrow{ \sf{ \: (lnx)(ln3) = (lny)(ln2)}}}$

we can see two unknown values here, but since we have to find the value of (x0), so we will take (lny) as subject to the equation,

${ \longrightarrow{ \sf{ \: (lny) = \frac{(lnx)(ln3)}{(ln2)} \: \: \: \: \: \: \: \: \: \: \: { \red{..(ii) }}}}}$

On putting the value of eq.(ii) in eq.(i),

${ \longrightarrow{ \green{\sf{ \: ( {ln)}^{2} - ( {ln3)}^{2} = (ln3){ \red{(lny)}} - (ln2)(lnx)}}}}$

Put value of (lny) here,

${ \longrightarrow{ \green{\sf{ \: {(ln)}^{2} - ( {ln3)}^{2} = (ln3) \: { \red{{ \huge{(}}{ \frac{(lnx)(ln3)}{(ln2)}{ \huge{)}} }}} - (ln2)(lnx) }}}}$

we can take (lnx) common on the right side,

${ \longrightarrow{ \green{\sf{ \: {(ln)}^{2} - ( {ln3)}^{2} = (ln3) \: { \red{{ \huge{(}}{ \frac{{ \blue{{ \underline{(lnx)}}}}(ln3)}{(ln2)}{ \huge{)}} }}} - (ln2){ \blue{ \underline{(lnx) }}}}}}}$

${ \longrightarrow{ \green{\sf{ \: ( {ln)}^{2} - ( {ln3)}^{2} = { \blue{(lnx)}} \: { \huge{(}} \frac{ {(ln3)}^{2} }{ln2} - ln2{ \huge{)}} }}}}$

${ \longrightarrow{ \green{\sf{ \: ( {ln)}^{2} - ( {ln3)}^{2} = (lnx){ \huge{(}} \frac{ {(ln3)}^{2} - {(ln2)}^{2} }{(ln2)} { \huge{)}}\: }}}}$

${ \longrightarrow{ \green{\sf{ \: ( {ln)}^{2} - ( {ln3)}^{2} = \frac{(lnx)}{(ln2)}{ \huge{(}} { {(ln3)}^{2} - {(ln2)}^{2} } { \huge{)}}\: }}}}$

${ \longrightarrow{ \green{\sf{ \: {\cancel{( {ln)}^{2} - ( {ln3)}^{2}}} = \frac{(lnx)}{(ln2)}{ \huge{(}} {{\cancel{ {(ln3)}^{2} - {(ln2)}^{2} }}} { \huge{)}}\: }}}}$

${ \longrightarrow{ \green{\sf{ \: -1 = \frac{(lnx)}{(ln2)}\: }}}}$

${ \longrightarrow{ \green{\sf{ \: -ln2 = lnx}}}}$

We can shift the (- sign) to its power,

${ \longrightarrow{ \green{\sf{ \: ln{(2 )}^{-1}= lnx}}}}$

Now, they have same base so equate the powers,

${ \longrightarrow{ \green{\sf{ \: {2}^{-1}= x}}}}$

${ \longrightarrow{\purple{\sf{ \: x = \frac{1}{2}}}}}$

Therefore,

• x0 = 1/2 (ans)