# Verifying solutions to logarithms

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?

• 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. Commented Sep 5, 2021 at 12:22

You have overlooked a simple fact

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

only works for $$00, ~n\in\mathbb R.$$

But remember that, if $$0, then you have

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

• 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? Commented Sep 6, 2021 at 12:26
• @Tom because, $$(-b)^{2k}=b^{2k}=|b|^{2k}$$ Commented Sep 6, 2021 at 12:34
• okay got it thanks Commented Sep 7, 2021 at 11:50