Question 261432
<pre><font size = 4 color = "indigo"><b>
{{{log2^y+log2(y+2)=log2^3}}} base is 2 for all

I think you meant

{{{log(2,(y))+log(2,(y+2))=log(2,(3))}}}

In algebra.com triple bracket notation, that's written

{{{"{{{"}}}{{{"log(2,(y))+log(2,(y+2))=log(2,(3))"}}}{{{"{{{"}}}  

Anyway,

{{{log(2,(y))+log(2,(y+2))=log(2,(3))}}}

Use the principle of logs:  {{{log(B,(A))+log(B,(C))=log(B,(AC))}}}
to rewrite the left side:

{{{log(2,(y(y+2)))=log(2,(3))}}}

Now use the principle that if {{{log(B,(A))=log(B,(C))}}} then{{{A=C}}}

{{{y(y+2)=3}}}
{{{y^2+2y=3}}}
{{{y^2+2y-3=0)}}}
{{{(y+3)(y-1)=0}}}
{{{y+3=0}}}; {{{y-1=0}}}
{{{y=-3}}};   {{{y=1}}}

Now we must dicard the answer {{{y=-3}}},
because if we were to substitute that into
the original equation, the first term would be
{{{log(2,(-3))}}} and logs of negative numbers
are not defined as real numbers.  So the only
solution is {{{y=1}}}

Edwin</pre>