2^(log(2x,x+2))+3^(log(2,x+3))=sqrt(-1-x)
Can someone help me with this equation I don't know how to start Any hint would be appreciated Thank you!
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If you mean $$2^{\log_{2x}(x+2)}+3^{\log_{2}(x+3)}=\sqrt{-1-x}$$ then the domain gives $x>0$ and $x\leq-1$, which is impossible.
Michael Rozenberg
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so we can say that 2x must be greater than 0 right? 2x>0? cause i had doubts about that? Does that mean that in any logarithmic equation we must write base as greater than zero ? – Apr 18 '17 at 10:53
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@John Doe Yes, you are right. If we write $\log_{a}b$ then $a>0$, $b>0$ and $a\neq1$ by definition. – Michael Rozenberg Apr 18 '17 at 10:55
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okay thank you very much :) – Apr 18 '17 at 11:08