Use log rules and solve for k

Question

T = 2π √(m/k)

Answer 1

Solving for k would not usually use log rules, but we can do it both ways to show differences:

Divide both sides by 2π: T/2π = √(m/k)

Take the log of both sides (usually we use natural log): ln (T/2π) = ln (√m/k)

Use log rule (treating the sq. root as an exponent = 1/2): ln (T/2π) = 1/2 ln (m/k)

Multiply by 2: 2ln (T/2π) = ln (m/k)

Use log rule for quotients: 2ln (T/2π) = ln (m) - ln (k)

Isolate ln(k): ln(k) = ln (m) - 2ln (T/2π)

Use log rules again on right (his time in reverse): ln(k) = ln (m/(T/2π)²)

Fraction rule: ln(k) = ln (4π²m/T²)

Take e to the power of both sides to get rid of lns: k = 4π²m/T²

That is incredibly cumbersome compared to just squaring:

T/2π = √(m/k)

T²/4π² = m/k

4π²/T² = k/m

k = 4π²m/T²