Question 475999
1.) {{{log((x+2), 16)=4}}}
Formula {{{log(c,a)=b}}} <=> {{{a=c^b}}}
{{{16=(x+2)^4}}}
We can look at 16 like at {{{2^4}}}
{{{2^4=(x+2)^4}}}
{{{x+2=2}}}
{{{x=2-2}}}
{{{x=0}}}
Check {{{log((0+2), 16)=4}}}
{{{log(2, 16)=4}}}
{{{log(2, 2^4)=4}}}
{{{4=4}}} correct
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2.) {{{log(3,x) + log(3,(x-2)) = 1}}}
{{{system(x>0,x-2>0)}}} => {{{system(x>0,x>2)}}} => {{{x>2}}}
Use formula {{{log(c,a)+log(c,b)=log(c, (a*b))
{{{log(3,(x(x-2)) )= 1}}}
{{{x(x-2)=3^1}}}
{{{x^2-2x-3=0}}}
 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
 {{{x = (-(-2)+- sqrt( (-2)^2-4*1*(-3) ))/(2*1) }}} 
 {{{x = (2 +- sqrt( 16))/2 }}} 
 {{{x = (2 - 4)/2 =-1<2}}} the extraneous root
 {{{x = (2 +4)/2 =3}}}
Answer: 3