Question

if log 2 to the base x+ log 8 the base x+ log 64 to the base xequal to 3 then x

Answer 1

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

Given that

$\rm :\longmapsto\: log_{x}(2) + log_{x}(8) + log_{x}(64) = 3$

$\rm \: \: \rm :\longmapsto\:\: \: log_{x}(2 \times 8 \times 64) = 3$

$\: \: \: \: \: \red{\bigg \{ \because \: log_{a}(x) + log_{a}(y)=log_{a}(xy)\bigg \}}$

$\rm \: \: \rm :\longmapsto\: \: \: log_{x}(2 \times {2}^{3} \times {2}^{6}) = 3$

$\rm \: \: \rm :\longmapsto\: \: \: log_{x}( {2}^{1 + 3 + 6} ) = 3$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {a}^{x} \times {a}^{y} = {a}^{x + y} \bigg \}}$

$\rm \: \: \rm :\longmapsto\: \: \: log_{x}( {2}^{10} ) = 3$

$\rm :\longmapsto\: {2}^{10} = {x}^{3}$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log_{x}(y) = z \: then \: y = {x}^{z} \bigg \}}$

$\rm :\longmapsto\: {x}^{3} = {2}^{9} \times 2$

$\bf\implies \:x = {2}^{3} \times {2}^{ \frac{1}{3} }$

$\bf\implies \:x = 8 \: \sqrt[3]{2}$

Additional Information :-

$\boxed{ \sf \: log_{x}(x) = 1 }$

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

$\boxed{ \sf \: {a}^{ log_{a}(x) } = x }$

$\boxed{ \sf \: {a}^{ ylog_{a}(x) } = {x}^{y} }$

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

$\boxed{ \sf \: {e}^{ log_{e}(x) } = x}$