
How do you use a calculator to evaluate the expression $\log 0.8$ to four decimal places? \[\]
Answer
462k+ 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.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Is Cellular respiration an Oxidation or Reduction class 11 chemistry CBSE

In electron dot structure the valence shell electrons class 11 chemistry CBSE

What is the Pitti Island famous for ABird Sanctuary class 11 social science CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
