Solving $\log_{10}(x+1)+\log_{10}(x-1)=3$

Question

Question:

$$\log_{10}(x+1)+\log_{10}(x-1)=3$$

workings:

$$\log_{10}(x+1)(x-1)=3 \\ \log_{10}(x-1)^2=3 \\ (x-1)^2=10^3 \\ (x-1)^2=1000 \\ (x-1)=\sqrt{1000} \\ (x-1)=10\sqrt{10} \\ x=10\sqrt{10}+1$$

Any glaring errors or is this correct?

Answer 1

Careful: it is $\log(x^2-1)$ and not $\log(x-1)^2$.

And you forgot that from $x^2 =a$ you got $x=\pm \sqrt{a}$

Finally, you should check if your solution is $>1$ (because of domain of $\log(x-1)$)

Answer 2

Well, we know that:

$$\log_\alpha\left(\beta\right)=\frac{\ln\left(\beta\right)}{\ln\left(\alpha\right)}\tag1$$

So, in your example we are trying to solve:

$$\log_\alpha\left(x+1\right)+\log_\alpha\left(x-1\right)=\text{n}\tag2$$

Now, we rewrite:

$$\frac{\ln\left(x+1\right)}{\ln\left(10\right)}+\frac{\ln\left(x-1\right)}{\ln\left(10\right)}=\frac{\ln\left(x+1\right)+\ln\left(x-1\right)}{\ln\left(10\right)}=\text{n}\tag3$$

Using the $\ln\left(\alpha\right)+\ln\left(\beta\right)=\ln\left(\alpha\cdot\beta\right)$ law of logs, we can write:

$$\frac{\ln\left(\left(x+1\right)\cdot\left(x-1\right)\right)}{\ln\left(10\right)}=\frac{\ln\left(x^2-1\right)}{\ln\left(10\right)}=\text{n}\space\Longleftrightarrow\space\text{n}\cdot\ln\left(10\right)=\ln\left(x^2-1\right)\tag4$$

Using the $\beta\cdot\ln\left(\alpha\right)=\ln\left(\alpha^\beta\right)$ law of logs, we can write:

$$\ln\left(10^\text{n}\right)=\ln\left(x^2-1\right)\space\Longleftrightarrow\space10^\text{n}=x^2-1\space\Longleftrightarrow\space x=\pm\sqrt{1+10^\text{n}}\tag5$$

Now, using the correct domain we know that $x=\sqrt{1+10^\text{n}}$ is the correct solution.