\(\log _{10}(x)\) is usually written as \(\log (x)\). The natural logarithm of a real number \(x\) is \(\log _{e}(x)\) and is usually written as \(\ln (x)\).
In this post we are going to solve questions related to logarithm using its properties ( click here to check its properties)
1) QUESTION - Using property \(\;\; a^{c}=b\;\; \) if and only if \(\;\;\log_{a}b=c\;\;\)Write the following exponential form in logarithm.
i) \(\;2^4=16\)
SOLUTION-
\(2^4=16\)
\(\Rightarrow \log_{2}16=4\)
ii) \(\;5^{-3} = \displaystyle \frac{1}{125}\)
SOLUTION-
\(\;5^{-3} = \displaystyle \frac{1}{125}\)
\(\Rightarrow \displaystyle \log_{5} \frac{1}{125} = -3\)
iii) \(\;10^{-2} = 0.01 \)
SOLUTION-
\(\;10^{-2} =0.01\)
\(\Rightarrow \displaystyle \log_{10}0.01 = -2\)
iv) \(\;8^{2/3} = 4 \)
SOLUTION-
\(\;8^{2/3} =4\)
\(\Rightarrow \displaystyle \log_{8} 4= \frac{2}{3}\)
Now we will check how to find the value of a given Logarithm form .
i) \(\;\log_{3} 81 \)
SOLUTION-
\(\;\log_{3}81\)
let say we don't know its value, take it \(\;x\)
\(\Rightarrow \log_{3} 81= x \)
From above we know that \(\;\; \log_{a} b = n \;\; \Rightarrow a^n =b\;\) by using this above can be written as
\(\Rightarrow 3^{x}=81\)
now we have to think what power of 3 should we raise so that it will give 81, and the answer is 4
\(\Rightarrow 3^{x} =3^{4}\)
\(\Rightarrow x=4\)
There is also an alternate way , but for that first check these indenitites (CLICK HERE).
ANOTHER WAY-
\(\log_{3} 81 = \log_{3} (3^4)\)
\(\Rightarrow \;\;\;\; \log_{3} 81 = 4 \log_{3} (3)\cdots\;\;(1)\)
Because we have identity that \(\;\; \log_{a}(b)^n = n \log_{a} (b)\)
so form (1) we have
\(\log_{3} 81 = 4 \log_{3} (3)\cdots\)
again using property of Logarithm that \(\;\log_{a}(a) = 1\;\;\) we will get
\(\log_{3} 81 = 4(1)\)
ANSWER - 4
SOLUTION-
Lets say its value is \(\;x\;\)
then we will have \(\; \log_{\frac{1}{5}} (625) =x\)
\(\Rightarrow \; \displaystyle \left(\displaystyle \frac{1}{5}\right)^{x}=625\)
So now we have to look for \(\;x\;\) which satisfies above condition.
\(\; \displaystyle \left(\frac{1}{5}\right)^{x}=625\)
\(\; \displaystyle \left(\displaystyle \frac{1}{5}\right)^{x}= 5^4\)
\(\; \displaystyle \left(\frac{1}{5}\right)^{x}= \left(\frac{1}{5}\right)^{-4}\)
from here this is easy to say that \(\;x=-4\)
ANSWER - \(\;-4\;\)
v). \(\displaystyle \; \log_{2}\left( 3^{-\log_{3}(2)}\right)\)
iii). \(\; \displaystyle \log \left (\frac{1}{1000} \right)\)
SOLUTION- Whenever the base of logarithm is not given consider that its base is 10.
NOTE - Generally books use \(\; \log \;\) instead of \(\; \log_{10}\;\) and \(\; \ln\;\) for \(\; \log_{e}\;\).
\(\; \displaystyle \log \left (\frac{1}{1000} \right) =x\)
\( \Rightarrow \displaystyle \; 10^{x} = \left (\frac{1}{1000} \right)\)
or
\(\Rightarrow \displaystyle\; 10^{x} = \left (\frac{1}{10} \right)^3\)
\(\Rightarrow \displaystyle \; 10^{x} = \left (\frac{10}{1} \right)^{-3}\)
\(\Rightarrow \displaystyle \; 10^{x} = \left (10^{-3}\right)\)
\(\Rightarrow \;x = -3\;\)
SECOND WAY
Given that \(\; \displaystyle \log \left (\frac{1}{1000} \right) \cdots \;\;\;\;(2) \)
we will use identity \(\;\log_{a} \left(\displaystyle \frac{b}{c}\right) =\log_{a}(b) -\log_{a}(c) \)
so using above property in (2),
we will get
\(\; \displaystyle \log \left (\frac{1}{1000} \right) =\log(1) -\log(1000) \)
\(\Rightarrow \displaystyle \log \left (\frac{1}{1000} \right) =0 -\log(10)^{3} \)
using property that \(\;\log_{a} (b)^n = n\log_{a} (b)\)
\(\Rightarrow \displaystyle \log \left (\frac{1}{1000} \right) = -3\log(10) \)
using identity that \(\log_{a} (a)=1\;\; \text{we can write } \;\log(10) =1\)
\(\Rightarrow \displaystyle \log \left (\frac{1}{1000} \right) =-3(1) \)
ANSWER : -3
iv ) \(\; \log \left( \sqrt[3]{10^{5}}\right)\)
SOLUTION
\(\; \log \left( \sqrt[3]{10^{5}}\right)= \; \log \left( 10^{5/3}\right) \)
\(\Rightarrow \; \displaystyle \frac{5}{3}\log \left( 10\right)\)
\(\Rightarrow \; \displaystyle \frac{5}{3}\)
ANSWER : \(\;\displaystyle \frac{5}{3}\)
v). \(\displaystyle \; \log_{2}\left( 3^{-\log_{3}(2)}\right)\)
SOLUTION:
\(\displaystyle \log_{2} \left( 3^{-\log_{3}(2)}\right) =\displaystyle \; \log_{2}\left( 3^{\log_{3}(2)^{-1}}\right) \)
\(\Rightarrow \displaystyle \; \log_{2}\left( 3^{\log_{3}\left(\frac{1}{2}\right)}\right)\)
We have a property of logrithm that, \(\; a^{\log_{a} b } =b\;\) we will use this is above
\(\Rightarrow \displaystyle \; \log_{2}\left(\frac{1}{2}\right)\)
\(\Rightarrow \displaystyle \; \log_{2}\left(\frac{1}{2}\right)\)
\(\Rightarrow \displaystyle \; \log_{2}\left({2}^{-1}\right)\)
\(\;=-1\)
ANSWER : -1
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