# Solve for $x\quad \log_2(2^n) = \log_2(1+x)$

I am out of practice with logs, but this is derived from the channel capacity theorem.

$$B\log_2\left(1 + \frac SN\right)$$

Solve for $x$

$$\log_2(2^n) = \log_2(1+x)$$

I need this equation manipulated so that $x$ is the answer.

thanks!!!

• HINT: $\log_kx=\log_ky \iff x=y$. Feb 4 '16 at 16:20

$$\log_2(2^n)=\log_2(1+x)\Longrightarrow 2^n=1+x\Longrightarrow x=2^n-1$$

• Thank You!!! to expand n=8 ; 2^8-1 = 255 so S/N = 255; 10Log(255) = 24.07dB Feb 4 '16 at 16:15

Notice, a few things about logs:

• $$\log_a(x)=\frac{\ln(x)}{\ln(a)}$$
• $$\ln(e)=\log_e(e)=\frac{\ln(e)}{\ln(e)}=1$$
• $$\exp(\ln(x))=e^{\ln(x)}=x$$
• $$\ln(x)=\log_e(x)=\frac{\ln(x)}{\ln(e)}=\frac{\ln(x)}{1}=\ln(x)$$
• $$\ln(a^x)=x\ln(a)\space\space\space\text{when}\space a,x\space\text{are positive}$$
• $$\ln\left(\frac{a}{x}\right)=\ln(a)-\ln(x)\space\space\space\text{when}\space a,x\space\text{are positive}$$

$$\log_2\left(2^n\right)=\log_2\left(1+x\right)\Longleftrightarrow$$ $$\frac{\ln\left(2^n\right)}{\ln(2)}=\frac{\ln\left(1+x\right)}{\ln(2)}\Longleftrightarrow$$ $$\ln(2)\ln\left(1+x\right)=\ln(2)\ln\left(2^n\right)\Longleftrightarrow$$ $$\ln\left(1+x\right)=\ln\left(2^n\right)\Longleftrightarrow$$ $$e^{\ln\left(1+x\right)}=e^{\ln\left(2^n\right)}\Longleftrightarrow$$ $$1+x=2^n\Longleftrightarrow$$ $$x=2^n-1$$