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log2(x) + log2(x+3) = log2(2)
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\[\log_{2} (x*(x+3)) = \log_{2} (2)\rightarrow \log_{2} (x ^{2}+3x) = 1\]
how do i find "x"?
a good rule to know is \[n^{\log_{n} (x)} = x\] where n can be any number (in your case it is 2)
\[2^{1}=x ^{2}+3x\]
You know that log(x) + log(y) = log(xy) so, your equation is equivalent to log2(x(x+3)) = log2(2) which means that x(x+3) = 2 iff x^2+3x-2=0 which can be calculated using delta
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\[ x ^{2}+3x -2 = 0\]
this you should now know what to do.
thank you
you're welcome
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