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I was working on a logarithm question and I know how to solve it.

$\log_2(y^2) = 4+\log_2(y+5)$

Now after working out the solution you will find that the quadratic yields

$y=-4, y=20$

now you are supposed to verify your solutions to rid of any that do not comply with logs requirements for

$\log_a(b)$

$a,b>0$

log laws state that

$\log_a(b^c)= (c)\log_a(b)$

from the equation above

$\log_2(y^2) = 4+\log_2(y+5)$

both solutions work

however when you pull out the squared

$2\log_2(y) = 4+\log_2(y+5)$

since $y=-4$ is not positive, it is not valid.

but since $2\log_2(y)$ and $\log_2(y^2)$ are equivalent what is the problem here with depending on which form is in the equation, the solutions can be different?

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  • $\begingroup$ Obviously since $2\log_2(y)$ is only defined for positive $y$ and $\log_2(y^2)$ is defined for all non-zero $y$ they aren't equivalent. However to keep the awesome property we can look at $y^2=|y|^2$, then both $2\log_2(|y|)$ and $\log_2(y^2)$ are defined for all non zero $y$ and are equivalent. $\endgroup$
    – kingW3
    Commented Sep 5, 2021 at 12:22

1 Answer 1

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You have overlooked a simple fact

$$\log_ab^n=n\log_ab$$

only works for $0<a≠1,~b>0, ~n\in\mathbb R.$

But remember that, if $0<a≠1,~b<0,~n=2k, ~k\in\mathbb Z^{+}$, then you have

$$\log_ab^n=n\log_a|b|.$$

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  • $\begingroup$ oh okay I see there's actually some more restrictions on using the power rule. And from what I am seeing, the bottom one works due to the base being even. but why exactly is there an absolute value? $\endgroup$
    – Tom Xia
    Commented Sep 6, 2021 at 12:26
  • $\begingroup$ @Tom because, $$(-b)^{2k}=b^{2k}=|b|^{2k}$$ $\endgroup$
    – lone student
    Commented Sep 6, 2021 at 12:34
  • $\begingroup$ okay got it thanks $\endgroup$
    – Tom Xia
    Commented Sep 7, 2021 at 11:50

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