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# How do you simplify ${{\log }_{5}}\left[ \dfrac{1}{250} \right]$ ?

Last updated date: 20th Jun 2024
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Hint: Here, in this question we have to simplify the given function or you can write the function in another way. After that we must use the $\log \left( xy \right)=\log x+\log y$ by putting the value in this we get the value. Also we can use $\log \left( \dfrac{x}{y} \right)=\log \left( x \right)-\log \left( y \right)$ to simplify it, though we get the same answer. We will consider both. By following the given condition we have to solve the problem.

Complete step by step solution:
We have,
${{\log }_{5}}\left[ \dfrac{1}{250} \right]$
We can write the given $\log$ as,
${{\log }_{5}}\left[ \dfrac{1}{250} \right]={{\log }_{5}}{{\left( 250 \right)}^{-1}}=-{{\log }_{5}}\left( 250 \right)$
We will use the formula of $\log \left( xy \right)$
The equation we get,
$\log \left( xy \right)=\log x+\log y$
$\therefore -\log \left( 2\times 125 \right)=-\left[ {{\log }_{5}}\left( 2 \right)+{{\log }_{5}}\left( 125 \right) \right]$
Then by using definition, we get
$=-{{\log }_{5}}\left( 2 \right)-3$
Or we alternatively other formula we use,
$\log \left( \dfrac{x}{y} \right)=\log \left( x \right)-\log \left( y \right)$
$\therefore {{\log }_{5}}\left[ \dfrac{1}{250} \right]={{\log }_{5}}\left[ 1 \right]-{{\log }_{5}}\left[ 250 \right]$
$={{\log }_{5}}\left[ 1 \right]-{{\log }_{5}}\left[ 2\times 125 \right]$
$\therefore {{\log }_{5}}\left[ \dfrac{1}{250} \right]={{\log }_{5}}\left[ 1 \right]-\left[ {{\log }_{5}}\left[ 2 \right]+{{\log }_{5}}\left[ 125 \right] \right]$
Again use the definition of $\log$ we get,
$=0-{{\log }_{5}}\left( 2 \right)-3$
$=-{{\log }_{5}}\left( 2 \right)-3$

Hence,
After simplifying the ${{\log }_{5}}\left[ \dfrac{1}{250} \right]$ we get
The value as $-{{\log }_{5}}\left( 2 \right)-3$

In any exponential function and logarithm any number can be the base. However there are two bases used and also there are special names for these logarithms. These are the common and natural logarithms. Common logarithm (base $10$ written $\log x$ with a base) and natural logarithm (base $e,$ written$\ln x$) are also used. Scientific and graphing calculates have keys or Manu items that allows you easily find $\log x$ and$\ln x$, as well as $10x$ and ${{e}^{x}}.$ By using these given keys and change the base formula you will determine the logarithms in any base. Sometimes the input for the logarithm or exponent on the input for the logarithm or exponent on the base is more complicated than just a single variable. In this case you used the correct variable. In this case you used a single variable. In this case you used the correct input on the calculator. If you calculate the input last method for logarithm or calculate the input separately or can use it and then write it down.
As an example we take when we have to calculate $\log \left( 3x \right)$ where we have $x=4$ the answer we get $1.079.$ but if you didn’t use the bracket that is $\log 3x$ then the calculator find $\log 3$and multiply it to $4$ then we get $1.90.$ Which is the incorrect answer.
Note: We have the option in the calculator, so use the calculators to determine the logarithms as a power base. You can also determine the logarithms to bases other than $e$ or $10$ by using change of base formula. We have the definition of a logarithm that is ${{\log }_{b}}\left( a \right)=C\Leftrightarrow {{b}^{c}}=a$ where $'b'$ is base, $'c'$ is the exponent and $'a'$ is argument. When we rewrite an exponential equation in $\log$ form in exponential form. It is very helpful for us to remember that base logarithm is the same as base of exponent.
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