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Applications of Logarithmic Functions

Compound Interest

A=P(1+in)nt\boxed{A=P\Bigg(1+\dfrac{i}{n}\Bigg)^{nt}} A=Final Amount ($)P=Initial Amount ($)i=Interest raten=Compound periodst=Time elapsed\begin{array}{rcl} A&=&\text{Final Amount (\$)}\\\\ P&=&\text{Initial Amount (\$)}\\\\ i&=&\text{Interest rate}\\\\ n&=&\text{Compound periods}\\\\ t&=&\text{Time elapsed} \end{array}

Richter Scale

R=log(II0)\boxed{R=\log\Bigg(\dfrac{I}{I_0}\Bigg)} R=Richter scale magnitudeI=IntensityI0=Threshold Intensity\begin{array}{rcl} R&=&\text{Richter scale magnitude}\\\\ I&=&\text{Intensity}\\\\ I_0&=&\text{Threshold Intensity} \end{array}

ph Scale

pH=logH+\boxed{\text{pH}=-\log_{}\text{H}^+} H+=Hydrogen Ion Concentrationph=ph unit\begin{array}{rcl} \text{H}^+&=&\text{Hydrogen Ion Concentration}\\\\ \text{ph}&=&\text{ph unit} \end{array}

Decibel Scale

dB=10log(II0)\boxed{dB=10\log\Bigg(\dfrac{I}{I_0}\Bigg)} dB=Decibel unitI=Final SoundI0=Threshold Sound\begin{array}{rcl} dB&=&\text{Decibel unit}\\\\ I&=&\text{Final Sound}\\\\ I_0&=&\text{Threshold Sound} \end{array}

Example: Applications of Logarithmic Functions

Mateo has $4500 to invest and is deciding between two options:
  • Option A: invest in an account that gives 3.25% compounded bi-weekly (every 2 weeks)
  • Option B: invest in an account that gives 3.20% compounded semi-monthly (twice a month)
How long will it take to double his investment with each option? Which is the better investment?

Option A: Option B:

P=4500i=0.0325n=26\begin{array}{rcl} P&=&4500\\\\ i&=&0.0325\\\\ n&=&26 \end{array} P=4500i=0.032n=24\begin{array}{rcl} P&=&4500\\\\ i&=&0.032\\\\ n&=&24 \end{array}

A=P(1+in)nt9000=4500(1+0.032526)26t2=(1.00125)26tlog2=26tlog1.00125t=log226log1.00125t21.34 years\begin{array}{rcl} A&=&P\Bigg(1+\dfrac{i}{n}\Bigg)^{nt}\\\\ 9000&=&4500\Bigg(1+\dfrac{0.0325}{26}\Bigg)^{26t}\\\\ 2&=&(1.00125)^{26t}\\\\ \log2&=&26t\log1.00125\\\\ t&=&\dfrac{\log2}{26\log1.00125}\\\\ t&\approx&21.34~\text{years} \end{array} A=P(1+in)nt9000=4500(1+0.03224)24t2=(1.00133)24tlog2=24tlog1.00133t=log224log1.00133t21.73 years\begin{array}{rcl} A&=&P\Bigg(1+\dfrac{i}{n}\Bigg)^{nt}\\\\ 9000&=&4500\Bigg(1+\dfrac{0.032}{24}\Bigg)^{24t}\\\\ 2&=&(1.00133)^{24t}\\\\ \log2&=&24t\log1.00133\\\\ t&=&\dfrac{\log2}{24\log1.00133}\\\\ t&\approx&21.73~\text{years} \end{array}

Therefore, Option A is the better option.

Practice: Applications of Logarithmic Functions

The hydrogen ion concentration of a certain liquid tests to be 0.0005. Is this liquid acidic or alkaline?

Practice: Applications of Logarithmic Functions

George has $4000 to invest into an RRSP account that gives 12.5% annually. In 30 years, George plans on withdrawing his savings to use for his retirement. Unfortunately, when he withdraws his savings, he gets taxed at
  • 10% for withdrawals up to $5000
  • 20% for withdrawals between $5000 and $15000
  • 30% for withdrawals over $15000.



Practice: Applications of Logarithmic Functions

Edmund and Gracie and tracking a substance and its half-life. They began with 300mg of the substance and after 15 days, they were left with 10mg. What is the half-life of this substance?
Extra Practice