Question

\(\mathrm{s}=30\left(1-\mathrm{e}^{\mathrm{ln}}\right)\) \(5=30\left(\mathrm{l}-\mathrm{e}^{k}\right)\) \(\mathrm{e}^{k}=\frac{5}{6}\) \(\mathrm{k}=\ln (5 / 6)\)

Short answer

#tag_title# Step 7: Summarize the solution #tag_content# We have solved the given equations and found the variables as follows: \(s = 30(1 - A)\) \(\mathrm{l} = 1\) \(k = \ln \left(\frac{5}{6}\right)\) These are the values of the variables for the given system of equations.

Step-by-step solution

Rewrite the first equation

Rewrite the equation \(s = 30\left(1-\mathrm{e}^{\mathrm{ln}}\right)\) as \(s = 30(1 - e^{\ln{A}})\).

Simplify using logarithmic properties

Recall the property of logarithms: \(e^{\ln{A}} = A\). Using this property, we can simplify the first equation: \(s = 30(1 - A)\)

Address the other system of equations

Now, let's focus on the system of equations: \(5 = 30\left(\mathrm{l}-\mathrm{e}^{k}\right)\) \(\mathrm{e}^{k} = \frac{5}{6}\) \(\mathrm{k} = \ln (5 / 6)\)

Solve for \(k\)

The third equation shows that \(e^k = \frac{5}{6}\). The fourth equation has \(k\) expressed as a logarithm. So the fourth equation gives us: \(k = \ln \left(\frac{5}{6}\right)\)

Connect \(k\) and the first system of equations

Since we have found \(k\) in terms of a logarithm, we can substitute the value of \(k\) into the second equation: \(5 = 30\left(\mathrm{l}-\mathrm{e}^{\ln \left(\frac{5}{6}\right)}\right)\) Then, using the property of logarithms we used in step 2, we get: \(5 = 30\left(\mathrm{l} - \frac{5}{6}\right)\)

Solve for \(\mathrm{l}\)

Now, we just need to solve for the variable \(\mathrm{l}\): \(5 = 30\left(\mathrm{l}-\frac{5}{6}\right)\) \(\frac{5}{30} = \mathrm{l}-\frac{5}{6}\) \(\frac{1}{6} = \mathrm{l}-\frac{5}{6}\) \(\mathrm{l} = \frac{1}{6}+\frac{5}{6} = 1\) So, we have found the variable \(\mathrm{l} = 1\).