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$$y = \log_2(y^{-1} + 4y)$$

How can I deal with the $y^{-1}$ and $4y$, also does identify mean find the domain, range and symmetry?

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  • $\begingroup$ Is $\log[2]$ supposed to be $\log_2$, i.e. base 2 logarithm? And $y^{-1}$ means $1/y$, not the inverse of $y$, correct? $\endgroup$
    – JohnD
    Commented Dec 12, 2014 at 4:54
  • $\begingroup$ @JohnD I've edited it. $\endgroup$ Commented Dec 12, 2014 at 4:54
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    $\begingroup$ @method "Identify" depends on what you mean (or what the question means). ;) $\endgroup$ Commented Dec 12, 2014 at 4:54
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    $\begingroup$ No $x$'s, just $y$'s on both sides? $\endgroup$
    – JohnD
    Commented Dec 12, 2014 at 4:55
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    $\begingroup$ I am guessing it is supposed to be a function in y so f(y)= instead of y = $\endgroup$
    – method
    Commented Dec 12, 2014 at 7:13

2 Answers 2

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As given in the screenshot $$y = \log_2(y^{-1} + 4y)$$ cannot be considered as a function (I suppose that you noticed that, in the list, this is the only case where you have only $y$'s in both sides).

The only thing I could say is that $y$ is a constant which is the solution of equation $$f(y)=y -\log_2(y^{-1} + 4y)=0$$ which corresponds to the intersection of the two curves.

If you plot $f(y)$ as a function of $y$, you will notice that $f(y)=0$ for a value of $y$ close to $y=4$ (polishing the solution would lead to $y=4.03432$).

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I assumed that it was supposed to be a function in y instead and here's my solution $$f(y)=\log_2(y^{-1}+4y)$$ for this function to be real: $$y^{-1}+4y > 0$$ $$4y > \frac{1}{y}$$ $$y^2>\frac{1}{4}$$ therefore $$|y|>\frac{1}{2} OR -\frac{1}{2}<y<\frac{1}{2}$$

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    $\begingroup$ I agree with your interpretation, but note that $y$ must be positive (else $y^{-1}+4y$ is negative and its logarithm isn't real). That justifies multiplying both sides of the inequality by $y$ but then $y^2>1/4$ implies $y>1/2$. $\endgroup$ Commented Dec 12, 2014 at 11:27

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