+26387 dB = 20*Log(p1/p2) I need to solve for p1 what is the formula?
\(\begin{array}{rcll} dB &=& 20\cdot \log{ \left( \frac{p_1}{p_2} \right) }\\\\ dB &=& \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } \qquad &| \qquad 10^{()}\\\\ 10^{dB} &=&10^{ \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } } \\\\ 10^{dB} &=& \left( \frac{p_1}{p_2} \right)^{20} \qquad &| \qquad 1^{\frac{1}{20}}\\\\ 10^{\frac{dB}{20}} &=& \frac{p_1}{p_2} \qquad &| \qquad \cdot p_2 \\\\ p_2\cdot 10^{\frac{dB}{20}} &=& p_1\\\\ \boxed{~ \mathbf{ p_1 } \mathbf{=} \mathbf{p_2\cdot 10^{\frac{dB}{20}} } ~}\\\\ \end{array}\)
+6251 \(dB = 20 \log\left(\dfrac {p_1}{p_2}\right) \\ \dfrac {dB}{20} = \log\left(\dfrac {p_1}{p_2}\right) \\ \dfrac {dB}{20} = \log(p_1)-\log(p_2) \\ \log(p_1) = \dfrac {dB}{20} + \log(p_2) \\ p_1 = 10^{\left( \frac {dB}{20} + \log(p_2)\right)}\)
.+26387 dB = 20*Log(p1/p2) I need to solve for p1 what is the formula?
\(\begin{array}{rcll} dB &=& 20\cdot \log{ \left( \frac{p_1}{p_2} \right) }\\\\ dB &=& \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } \qquad &| \qquad 10^{()}\\\\ 10^{dB} &=&10^{ \log{ [ \left( \frac{p_1}{p_2} \right)^{20} ] } } \\\\ 10^{dB} &=& \left( \frac{p_1}{p_2} \right)^{20} \qquad &| \qquad 1^{\frac{1}{20}}\\\\ 10^{\frac{dB}{20}} &=& \frac{p_1}{p_2} \qquad &| \qquad \cdot p_2 \\\\ p_2\cdot 10^{\frac{dB}{20}} &=& p_1\\\\ \boxed{~ \mathbf{ p_1 } \mathbf{=} \mathbf{p_2\cdot 10^{\frac{dB}{20}} } ~}\\\\ \end{array}\)
