Question 942774

{{{3^x*10^(2x )= 4* 20^(x-2)}}}....use logarithm


{{{log(3^x*10^(2x ))= log(4* 20^(x-2))}}}


{{{log(3^x)+log(10^(2x ))= log(4)+log( 20^(x-2))}}}


{{{log(3^x)+log(10^(2x ))= log(4)+log( 20^x/20^2))}}}


{{{log(3^x)+log(10^(2x ))= log(4)+log( 20^x)-log(20^2))}}}


{{{log(3^x)+log(10^(2x ))= log(4)+log( 2^x*10^x)-log(20^2)) }}}


{{{log(3^x)+2log(10^(x ))= log(2^2)+log( 2^x)+log(2^x*5^x)-log(2^4*5^2))}}} 


{{{x*log(3)+2log(2^x)+2log(5^x)= 2log(2)+x*log( 2)+log(2^x)+log(5^x)-log(2^4)-log(5^2)) }}}


{{{x*log(3)+2x*log(2)+2x*log(5)= 2log(2)+x*log( 2)+x*log(2)+x*log(5)-4log(2)-2log(5)) }}}


{{{x*log(3)+2x*log(2)+2x*log(5)-2x*log( 2)-x*log(5)= 2log(2)-4log(2)-2log(5))}}} 

{{{x*log(3)+x*log(5)= -2log(2)-2log(5)}}} 


{{{x(log(3)+log(5))=-2(log(5) +log(2))}}} 


{{{x=-2(log(5) +log(2)) /(log(3)+log(5))}}}


{{{x=-2(0.69897+0.301)/(0.477+0.69897)}}}


{{{x=-2(0.99997)/(1.17597)}}}


{{{x=-1.99994/1.17597}}}


{{{x=-1.700672636206706}}}


{{{x=-1.701}}}


if you check it, you get

{{{0.0000612277=0.0000612277}}}