Question

if c=log 15 to the base 10, d= log 50 to the base 20 then show that (5-d)/(2(c-2d+cd+1)=log 40 to the base 9​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{c = log_{10} 15 \: \: and \: \: d = log_{20} 50 }$

TO PROVE

$\displaystyle \sf{ \frac{5 - d}{2(c - 2d + cd + 1)} = log_{9} 40 }$

EVALUATION

Here it is given that

$\displaystyle \sf{c = log_{10} 15 \: \: and \: \: d = log_{20} 50 }$

Now

$\displaystyle \sf{c = log_{10} 15 }$

$\displaystyle \sf{ \implies \: c = \frac{ log 15 }{ log 10 } }$

$\displaystyle \sf{ \implies \: c = \frac{ log 5 + log 3 }{1} }$

$\displaystyle \sf{ \implies \: c = log 2 + log 5 + log 3 - log 2 }$

$\displaystyle \sf{ \implies \: c = log 10 + log 3 - log 2 }$

$\displaystyle \sf{ \implies \: c = 1 + log 3 - log 2 } \: \: \: - - - (1)$

Again

$\displaystyle \sf{ \: d = log_{20} 50 }$

$\displaystyle \sf{ \implies \: d = \frac{ log 50}{ log 20} }$

$\displaystyle \sf{ \implies \: d = \frac{ log 10 + log 5}{ log 10 + log 2} }$

$\displaystyle \sf{ \implies \: d = \frac{ 1 + log 5}{ 1 + log 2} }$

$\displaystyle \sf{ \implies \: d = \frac{ 1 + log 5 + log 2 - log 2}{ 1 + log 2} }$

$\displaystyle \sf{ \implies \: d = \frac{ 1 + log 10 - log 2}{ 1 + log 2} }$

$\displaystyle \sf{ \implies \: d = \frac{ 1 + 1 - log 2}{ 1 + log 2} }$

$\displaystyle \sf{ \implies \: d = \frac{ 2 - log 2}{ 1 + log 2} }$

$\displaystyle \sf{ \implies \: (1 + log 2)d = 2 - log 2 }$

$\displaystyle \sf{ \implies \: log 2 \times (d + 1)= 2 - d }$

$\displaystyle \sf{ \implies \: log 2= \frac{2 - d}{d + 1} } \: \: - - - - (2)$

From Equation 1 we get

$\displaystyle \sf{ \: c = 1 + log 3 - log 2 }$

$\displaystyle \sf{ \implies \: log 3 = c - 1 + log 2 }$

$\displaystyle \sf{ \implies \: log 3 = c - 1 + \frac{2 - d}{d + 1} }$

$\displaystyle \sf{ \implies \: log 3 = \frac{cd + c - d - 1 + 2 - d}{d + 1} }$

$\displaystyle \sf{ \implies \: log 3 = \frac{cd + c - 2d +1}{d + 1} } \: \: - - (3)$

Now RHS

$\displaystyle \sf{ = log_{9} 40 }$

$\displaystyle \sf{ = \frac{ log 40}{log 9} }$

$\displaystyle \sf{ = \frac{ log 10 + log 4 }{2log 3} }$

$\displaystyle \sf{ = \frac{ 1+2 log 2 }{2log 3} }$

$\displaystyle \sf{ = \frac{ 1+2 \bigg( \frac{2 - d}{d + 1} \bigg)}{2 \bigg( \frac{cd + c - 2d +1}{d + 1} \bigg)} }$

$\displaystyle \sf{ = \frac{ \bigg( \frac{d + 1 + 4 - 2d}{d + 1} \bigg)}{2 \bigg( \frac{cd + c - 2d +1}{d + 1} \bigg)} }$

$\displaystyle \sf{ = \frac{ \bigg( \frac{5 - d}{d + 1} \bigg)}{2 \bigg( \frac{cd + c - 2d +1}{d + 1} \bigg)} }$

$\displaystyle \sf{ = \frac{5 - d}{2(c - 2d + cd + 1)} }$

= LHS

Thus we get

$\boxed{ \: \: \displaystyle \sf{ \frac{5 - d}{2(c - 2d + cd + 1)} = log_{9} 40 } \: \: }$

Hence proved

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