Question 1102526
<pre>

Learn how to change logarithm equations to and from 
equivalent exponential equations:

{{{log(B,(A))=C}}} and {{{A=B^C}}} are equivalent.

Therefore your first equation:

{{{log(3,(x-4y+5))=0}}} is equivalent to

{{{x-4y+5=3^0}}} and then
{{{x-4y+5=1}}}
{{{x=4y-4}}}

Your second equation:

{{{log(3,(x-1))-log(3,(y))=1}}}

Using a rule of logarithms, becomes

{{{log(3,((x-1)y^""))=1}}}, which is equivalent to

{{{(x-1)y=3^1}}} and then

{{{(x-1)y=3}}}

Substituting 4y-4 for x:

{{{(4y-4-1)y=3}}}

{{{(4y-5)y=3}}

{{{4y^2-5y=3}}

{{{4y^2-5y-3=0}}

{{{y = (-(-5) +- sqrt((-5)^2-4(4)(-3) ))/(2(4)) }}}

{{{y = (5 +- sqrt(25+48))/8 }}}

{{{y = (5 +- sqrt(73))/8 }}}

Substituting in {{{x=4y-4}}}

{{{x=4((5 +- sqrt(73))/8)-4}}}

{{{x=(5 +- sqrt(73))/2-4}}}

{{{x=(5 +- sqrt(73))/2-8/2}}}

{{{x=(5-8 +- sqrt(73))/2}}}

{{{x=(-3 +- sqrt(73))/2}}}

Since logs cannot be taken of negative numbers, and
the original second equation contains {{{log(3,(y))}}} 
and {{{log(3,(x-1))}}}, the only solutions are the
positive ones:

{{{x=(-3 + sqrt(73))/2}}} and {{{y = (5 + sqrt(73))/8 }}}

Edwin</pre>