How do I simplify these logs?

Question

$$a\log\left(3+x\right)\cdot\frac{1}{3+x}=\log\left(3+x\right)+0.5r\log\left(3+x\right)$$ How do I solve for $x$? Can I not just divide through by $\log(3+x)$?

Answer 1

The equation is only defined for $x> -3.$

Case 1: $x=-2.$ Then $\log(3+x)= \log 1=0,$ hence $x=-2$ is a solution.

Case 2: $x \ne -2$. Then $\log(3+x) \ne 0.$ It follows

$$a\frac{1}{3+x}=1+0.5r.$$

Can you proceed ?

Answer 2

Yes you can do this but you have to be careful. You have to exclude the case $\log(3+x)=0$, since you are not allowed to divide by zero. That means that $$a\cdot\frac{1}{3+x}=1+0.5r$$ if $\log(3+x)\neq0$. You should be able to sove for $x$ in the equation above.

For what $x$ is $\log(3+x)=0$? Since $\log(1)=0$, $x$ must be $-2$. And $$a\cdot0\cdot\frac{1}{3+x}=0+0.5r\cdot0$$ simplifies to $0=0$, which is obviously correct, so $-2$ is also a solution to your equation. $$\mathbb{L}=\left\{-2,\frac{a}{1+0.5r}-3\right\}$$

Addendum: It depends if you work with real or complex numbers. If you include complex numbers, $\mathbb{L}$ is correct. If you want $x$ to be real, you have to put further conditions on $r$ and $a$ so that $$\frac{a}{1+0.5r}-3>-3.$$

Answer 3

Don't be intimidated or get tripped up by the $\log\left(3+x\right)$. Try to let $z=\log\left(3+x\right)$. Then we have

$$az\cdot\frac{1}{3+x}=z+0.5rz$$

Hopefully this seems like something you've seen before.

Now before dividing by something in an equality, we have to consider that we're dividing by zero.

So we take cases $z=0$ and $z \ne 0$.

Case 1: For $z \ne 0$

$$a\cdot\frac{1}{3+x}=1+0.5r$$

Then

$$a\cdot\frac{1}{1+0.5r}=3+x$$

Case 2: For $z=0$:

$0=\log(3+x) \iff 3+x=1$

So, it seems we have 2 possible values for $x$ as solutions, and they come from 2 possible cases of values for $z$ (zero or non-zero).

Finally, we must have $3+x > 0$ so that the expression '$\log(3+x)$' is defined.