Question

Find the value of the following logarithms without using a calculator.(a) log319(b) log51(c) lne5(d) log0.00001

Answer 1

Part a)

For this part, we can use the following properties:

`\begin{gathered} (1)/(a^n)=a^(-n)\Rightarrow\text{ Property of the exponents} \\ \log _aa^x=x\Rightarrow\text{ Property of logarithms} \end{gathered}`

So, applying the above property of exponents, we have:

`\begin{gathered} (1)/(9)=(1)/(3\cdot3) \\ (1)/(9)=(1)/(3^2) \\ (1)/(9)=3^(-2) \end{gathered}`

Now, applying the above property of logarithms, we have:

`\begin{gathered} \log _3(1)/(9)=\log _33^(-2) \\ $$\boldsymbol{\log _3(1)/(9)=-2}$$ \end{gathered}`Part b)

For this part, we can apply the following property of logarithms:

`\log _a1=0`

Then, in this case, we have:

`\begin{gathered} a=5 \\ \log _a1=0 \\ \boldsymbol{\log _51=0} \end{gathered}`Part c)

For this part, we can apply the following property of logarithms:

`\ln e^x=x`

So, we have:

`\begin{gathered} x=5 \\ \ln e^x=x \\ $$\boldsymbol{\ln e}^{\boldsymbol{5}}\boldsymbol{=5}$$ \end{gathered}`Part d)

For this part, we can rewrite 0.00001 like this:

`\begin{gathered} 0.00001=(0.00001)/(1) \\ 0.00001=(0.00001\cdot100,000)/(1\cdot100,000) \\ 0.00001=(1)/(100,000) \\ 0.00001=(1)/(10\cdot10\cdot10\cdot10\cdot10) \\ 0.00001=(1)/(10^5) \\ 0.00001=10^(-5) \end{gathered}`

Now, applying the above property of logarithms, we have:

`\begin{gathered} a=10\text{ and }x=-5 \\ \log _aa^x=x \\ \log 0.00001=\log _(10)10^(-5) \\ $$\boldsymbol{\log 0.00001=-5}$$ \end{gathered}`