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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!!!

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  • $\begingroup$ HINT: $\log_kx=\log_ky \iff x=y$. $\endgroup$ Feb 4 '16 at 16:20
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$$\log_2(2^n)=\log_2(1+x)\Longrightarrow 2^n=1+x\Longrightarrow x=2^n-1$$

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  • $\begingroup$ Thank You!!! to expand n=8 ; 2^8-1 = 255 so S/N = 255; 10Log(255) = 24.07dB $\endgroup$
    – MoMan
    Feb 4 '16 at 16:15
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Notice, a few things about logs:

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

So, solving your question:

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

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