Question

Q10. Find the value of (49)^{1 - log (2 base 7)} + 5^{-log (4 base 5)}​

Answer 1

$\rm \longrightarrow {(49)}^{(1 - log_{7}2)} + {(5)}^{ - log_{5}4 } \\ \\ \rm \longrightarrow {(49)}^{(log_{7}7 - log_{7}2)} + {(5)}^{ - log_{5}4 } \\ \\ \rm \longrightarrow {(49)}^{log_{7} (\frac{7}{2} ) } + {(5)}^{ - log_{5}4 } \\ \\ \rm \longrightarrow {( {7}^{2} )}^{log_{7} (\frac{7}{2} ) } + {(5)}^{ - log_{5}4 } \\ \\ \rm \longrightarrow {( {7} )}^{2log_{7} (\frac{7}{2} ) } + {(5)}^{ - log_{5}4 } \\ \\ \rm \longrightarrow {( {7} )}^{log_{7} {(\frac{7}{2} )}^{2} } + {(5)}^{ log_{5} {(4)}^{ - 1} } \\ \\ \rm \longrightarrow {{ \bigg(\dfrac{7}{2} \bigg )}^{}}^{2log_{7} {7} } + {((4)^{ - 1})}^{ log_{5} 5 } \\ \\ \rm \longrightarrow {{ \bigg(\dfrac{7}{2} \bigg )}^{}}^{2 } + \frac{1}{4} \\ \\ \rm \longrightarrow \dfrac{49}{4} + \dfrac{1}{4} \\ \\ \rm \longrightarrow \dfrac{50}{4} \\ \\ \rm \longrightarrow \dfrac{25}{2}$

Additional Information :-

$\boxed{\boxed{\begin{minipage}{4.2cm}\circ\sf\:\ln 1=0\\\circ\sf\:\ln e=1\\\circ\sf\:\ln x=y\leftrightarrow e^y=x\\\circ\sf\:e^{\ln x}=x,\:x>0\\\circ\sf\:\ln(e^x)=x,\:x\in\mathbb{\large R}\\\circ\sf\:\ln(xy)=\ln x+\ln y\\\circ\sf\:\ln(x/y)=\ln x-\ln y\\\circ\sf\:\ln(x^r)=r\ln x\\\circ\sf\:\ln x=log_e\:x\end{minipage}}}$

Answer 2

Given Question :-

Evaluate:-

$\sf \: {49}^{(1 - log_{7}(2)) } + {5}^{ - log_{5}(4) }$

$\huge\underline{\sf{Solution-}}$

Identities used :-

$\boxed{ \blue{ \bf \: log_{x}(y) = \dfrac{ log(y) }{ log(x) } }}$

$\boxed{ \blue{ \bf \: {x}^{ log_{x}(y) } = y}}$

$\boxed{ \blue{ \bf \: log(x) - log(y) = log\dfrac{x}{y}}}$

$\boxed{ \blue{ \bf \: log( {x}^{y} ) = y log(x)}}$

Let's solve the problem now!!

Consider

$\rm :\longmapsto\:\: {49}^{(1 - log_{7}(2)) } + {5}^{ - log_{5}(4) }$

$\rm \: =\:\: \sf \: {49}^{ \bigg(1 - \dfrac{ log(2) }{ log(7) } \bigg)} + {5}^{log_{5} {4}^{ - 1}}$

$\rm \: =\:\: \sf \: {49}^{ \bigg(\dfrac{ log(7) - log(2) }{ log(7) } \bigg)} + {4}^{ - 1}$

$\rm \: = \: \: {7}^{2 log_{7} \bigg(\dfrac{7}{2} \bigg) } + \dfrac{1}{4}$

$\rm \: = \: \: {7}^{log_{7} \bigg(\dfrac{7}{2} \bigg)^{2} } + \dfrac{1}{4}$

$\rm \: = \: \: {\bigg(\dfrac{7}{2} \bigg) }^{2} + \dfrac{1}{4}$

$\rm \: = \: \: \dfrac{49}{4} + \dfrac{1}{4}$

$\rm \: = \: \: \dfrac{49 + 1}{4}$

$\rm \: = \: \: \dfrac{50}{4}$

$\rm \: = \: \: \dfrac{25}{2}$

Additional Information :-

$\boxed{ \blue{ \bf \:logx + logy = log(xy)}}$

$\boxed{ \blue{ \bf \: log(1) = 0}}$

$\boxed{ \blue{ \bf \: log( {10}^{x} ) = x}}$

$\boxed{ \blue{ \bf \: log_{x}(y) = \dfrac{1}{ log_{y}(x) }}}$