SOLUTION: {{{log2^y+log2(y+2)=log2^3}}} base is 2 for all

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Question 261432: log2%5Ey%2Blog2%28y%2B2%29=log2%5E3 base is 2 for all
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

log2%5Ey%2Blog2%28y%2B2%29=log2%5E3 base is 2 for all

I think you meant

log%282%2C%28y%29%29%2Blog%282%2C%28y%2B2%29%29=log%282%2C%283%29%29

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

%22%7B%7B%7B%22%22log%282%2C%28y%29%29%2Blog%282%2C%28y%2B2%29%29=log%282%2C%283%29%29%22%22%7B%7B%7B%22  

Anyway,

log%282%2C%28y%29%29%2Blog%282%2C%28y%2B2%29%29=log%282%2C%283%29%29

Use the principle of logs:  log%28B%2C%28A%29%29%2Blog%28B%2C%28C%29%29=log%28B%2C%28AC%29%29
to rewrite the left side:

log%282%2C%28y%28y%2B2%29%29%29=log%282%2C%283%29%29

Now use the principle that if log%28B%2C%28A%29%29=log%28B%2C%28C%29%29 thenA=C

y%28y%2B2%29=3
y%5E2%2B2y=3
y%5E2%2B2y-3=0%29
%28y%2B3%29%28y-1%29=0
y%2B3=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%282%2C%28-3%29%29 and logs of negative numbers
are not defined as real numbers.  So the only
solution is y=1

Edwin