Biology Forums - Study Force

\(-3.25=10\cdot \log \left(\frac{P_2}{2750}\right)\)

To isolate for \(P_2\), we need to divide both sides by the factor 10 before the log:

\(\frac{-3.25}{10}=\frac{10\cdot \log \left(\frac{P_2}{2750}\right)}{10}\)

\(\frac{-3.25}{10}=\log \left(\frac{P_2}{2750}\right)\)

Now, to remove the log so we can access the \(P_2\), we raise both sides of the equations as exponents to the base 10:

\(10^{\frac{-3.25}{10}}=10^{\log \left(\frac{P_2}{2750}\right)}\)

Because log is of base 10, by raising it as a power, it eliminates it:

\(10^{\frac{-3.25}{10}}=\frac{P_2}{2750}\)

One more step, multiply both sides by the denominator found on the right-side:

\(\left(2750\right)\cdot 10^{\frac{-3.25}{10}}=\frac{P_2}{2750}\cdot \left(2750\right)\)

\(\left(2750\right)\cdot 10^{\frac{-3.25}{10}}=P_2\)\

Therefore:

\(P_2=\left(2750\right)\cdot 10^{\frac{-3.25}{10}}=1301.16\)

Hope that helps!

Thank you!