Am I doing this right? $2 \log_2(x)- \log_2(1)-x=3$

Question

I know that the logairhtm with base anything of $1$ is $0$, which puts me at $2 \log_2(x)-x=3$. Also, I realize $\log_2(x)^2=2 \log_2(x)$. When it comes to the $x$ standing alone, I also said that $x=\log_2(2^x)$. Similarly, I know that $\log_2(8)=3$. However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.

Answer 1

However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.

As pointed out in the comments, it's no surprise you get stuck because the equation $$2 \log_2(x)- \log_2(1)-x=3$$ doesn't have a "nice" solution which you can easily find manually. If you just began learning about logarithms in an algebra-precalculus context, it's very unlikely that the equation was supposed to be like this.

So there is probably an error in the equation, or you miscopied or misinterpreted a part. My best guess is you (or the question setter) missed brackets and that it was supposed to be: $$2 \log_2(x)- \log_2\color{blue}{(1-x)}=3$$

Perhaps you can take it from there already and if not, here's a start:

$$\log_2(x^2)- \log_2\color{blue}{(1-x)}=3\iff\log_2\left(\frac{x^2}{1-x}\right)=3\iff \ldots$$

Answer 2

Well, solving a more general problem:

$$\text{n}\cdot\log_\text{n}\left(x\right)-\log_\text{n}\left(\text{n}-1\right)-x=1+\text{n}\tag1$$

Now, we know that:

• $$\log_\alpha\left(\beta\right)=\frac{\ln\left(\beta\right)}{\ln\left(\alpha\right)}\tag2$$
• $$\alpha\cdot\ln\left(\beta\right)=\ln\left(\beta^\alpha\right)\tag3$$
• $$\gamma=\log_\alpha\left(\alpha^\gamma\right)\tag4$$
• $$\ln\left(\alpha\right)+\ln\left(\beta\right)=\ln\left(\alpha\cdot\beta\right)\tag5$$
• $$\ln\left(\alpha\right)-\ln\left(\beta\right)=\ln\left(\frac{\alpha}{\beta}\right)\tag6$$

So, we can write:

$$\frac{\ln\left(x^\text{n}\right)}{\ln\left(\text{n}\right)}-\frac{\ln\left(\text{n}-1\right)}{\ln\left(\text{n}\right)}-\frac{\ln\left(\text{n}^x\right)}{\ln\left(\text{n}\right)}=\frac{\ln\left(\text{n}^{1+\text{n}}\right)}{\ln\left(\text{n}\right)}\tag7$$

Now, formula $(7)$ gives (when we assume $\ln\left(\text{n}\right)\ne0$):

$$\ln\left(x^\text{n}\right)-\ln\left(\text{n}-1\right)-\ln\left(\text{n}^x\right)=\ln\left(\text{n}^{1+\text{n}}\right)\tag8$$

Now, using rule $(5)$ and $(6)$ we can write:

$$\ln\left(\frac{x^\text{n}}{\text{n}^x\cdot\left(\text{n}-1\right)}\right)=\ln\left(\text{n}^{1+\text{n}}\right)\space\Longleftrightarrow\space\frac{x^\text{n}}{\text{n}^x\cdot\left(\text{n}-1\right)}=\text{n}^{1+\text{n}}\space\Longleftrightarrow\space$$ $$\frac{x^\text{n}}{\text{n}^x}=\text{n}^{1+\text{n}}\cdot\left(\text{n}-1\right)\tag9$$