How do you use a calculator to evaluate the expression $\log 0.8$ to four decimal places? \[\]
Answer
608.7k+ views
Hint: We recall the definition of logarithm, argument of logarithm and the common logarithm. In order to find the value of $\log 0.8$ we need to type the $\log $l button in the scientific calculator and put the value of the argument in the bracket. \[\]
Complete answer:
We know that the logarithm is the inverse operation to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$ must be raised, to produce that number$x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
Here $x$ is called the argument of the logarithm which is always positive. If the base $b=10$ we call logarithm common logarithm and write without base as $\log x$.
We are asked to evaluate the value of $\log 0.8$ using a calculator. We can find logarithm only scientific or graphic calculators. \[\]
Step-1:We turn on the calculator. \[\]
Step-2:We find the $\log $button somewhere in the top or middle. We press it will show$\log \left( {} \right)$ or $\log \left( {} \right.$. \[\]
Step-3: We type $0.8$ in the bracket as $\log \left( 0.8 \right)$. If there is only one bracket we type $\log \left( 0.8 \right.$ and then close the bracket $\log \left( 0.8 \right)$.\[\]
Step-4:We press $\text{= or EXE}$ to display the value as $-0.0\text{9691}00\text{13}$\[\]
So the value of $\log 0.8$ up to 4 decimal is $-0.0969$
Note: We can also find $\log 0.8$ without a calculator with logarithmic identities and the known value $\log 2=0.30102$. Let us consider
\[\log 0.8=\log \dfrac{8}{10}\]
We use the identity of quotient $\log \left( \dfrac{m}{m} \right)=\log m-\log n$ for $m=8,n=10$ in the above step to have;
\[\begin{align}
& \Rightarrow \log 0.8=\log 8-\log 10 \\
& \Rightarrow \log 0.8=\log {{2}^{3}}-\log 10 \\
\end{align}\]
We use identity of power $\log {{x}^{m}}=m\log x$ for $x=2.m=3$ and the known value $\log 10=1$ in the above step to have ;
\[\Rightarrow \log 0.8=3\log 2-1\]
We put the known logarithmic value $\log 2=0.30102$ in the above step to have;
\[\Rightarrow \log 0.8=3\times 0.301029-1=0.90306-1=0.0964\]
We can improve the values of $\log 2$ to find with more accuracy.
Complete answer:
We know that the logarithm is the inverse operation to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$ must be raised, to produce that number$x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
Here $x$ is called the argument of the logarithm which is always positive. If the base $b=10$ we call logarithm common logarithm and write without base as $\log x$.
We are asked to evaluate the value of $\log 0.8$ using a calculator. We can find logarithm only scientific or graphic calculators. \[\]
Step-1:We turn on the calculator. \[\]
Step-2:We find the $\log $button somewhere in the top or middle. We press it will show$\log \left( {} \right)$ or $\log \left( {} \right.$. \[\]
Step-3: We type $0.8$ in the bracket as $\log \left( 0.8 \right)$. If there is only one bracket we type $\log \left( 0.8 \right.$ and then close the bracket $\log \left( 0.8 \right)$.\[\]
Step-4:We press $\text{= or EXE}$ to display the value as $-0.0\text{9691}00\text{13}$\[\]
So the value of $\log 0.8$ up to 4 decimal is $-0.0969$
Note: We can also find $\log 0.8$ without a calculator with logarithmic identities and the known value $\log 2=0.30102$. Let us consider
\[\log 0.8=\log \dfrac{8}{10}\]
We use the identity of quotient $\log \left( \dfrac{m}{m} \right)=\log m-\log n$ for $m=8,n=10$ in the above step to have;
\[\begin{align}
& \Rightarrow \log 0.8=\log 8-\log 10 \\
& \Rightarrow \log 0.8=\log {{2}^{3}}-\log 10 \\
\end{align}\]
We use identity of power $\log {{x}^{m}}=m\log x$ for $x=2.m=3$ and the known value $\log 10=1$ in the above step to have ;
\[\Rightarrow \log 0.8=3\log 2-1\]
We put the known logarithmic value $\log 2=0.30102$ in the above step to have;
\[\Rightarrow \log 0.8=3\times 0.301029-1=0.90306-1=0.0964\]
We can improve the values of $\log 2$ to find with more accuracy.
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