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# The product of ${{\log }_{2}}17,{{\log }_{15}}2,{{\log }_{3}}\dfrac{1}{5}$ lies between two successive integers which are _______ and _______. 

Last updated date: 07th Sep 2024
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Hint: We change the base for logarithms involved in the given question to new base $d=10$ using the base change formula ${{\log }_{b}}x=\dfrac{{{\log }_{d}}x}{{{\log }_{d}}b}$ and then multiply. We use logarithmic identity of product ${{\log }_{b}}mn={{\log }_{b}}m+{{\log }_{n}}n$ and logarithmic identity of quotient ${{\log }_{b}}\dfrac{m}{n}={{\log }_{b}}m-{{\log }_{b}}n$ to proceed and use the logarithmic table to evaluate.

Complete step-by-step solution
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 $x,y,b$ are real numbers subjected to the condition that the argument of logarithm $x$ is always a positive number and $b$ is a positive number excluding 1. If we want to change the base of the logarithms to new base say $d>0,d\ne 1$ then we can do it using following formula,
${{\log }_{b}}x=\dfrac{{{\log }_{d}}x}{{{\log }_{d}}b}$
We multiply the three given logarithms in the question and have
${{\log }_{2}}17\times {{\log }_{15}}2\times {{\log }_{3}}\dfrac{1}{5}$
We change the base of each logarithm to base $d=10$ using the base change formula with $x=17,b=2$ in the first logarithm, with $x=2,b=15$ in the second logarithm and with $x=\dfrac{1}{5},b=3$ in third logarithm. We have,
\begin{align} & \Rightarrow {{\log }_{2}}17\times {{\log }_{15}}2\times {{\log }_{3}}\dfrac{1}{5} \\ & \Rightarrow \dfrac{\log 17}{\log 2}\times \dfrac{\log 2}{\log 15}\times \dfrac{\log \dfrac{1}{5}}{\log 3} \\ & \Rightarrow \dfrac{\log 17}{\log 15}\times \dfrac{\log \dfrac{1}{5}}{\log 3} \\ & \Rightarrow \dfrac{\log 17}{\log \left( 3\times 5 \right)}\times \dfrac{\log \dfrac{1}{5}}{\log 3} \\ \end{align}
We use the logarithmic identity of product ${{\log }_{b}}mn={{\log }_{b}}m+{{\log }_{n}}n$ for $m=3,n=5,b=10$and logarithmic identity of quotient ${{\log }_{b}}\dfrac{m}{n}={{\log }_{b}}m-{{\log }_{b}}n$ for $m=1,n=5,b=10$ in the above step and proceed to have,
$\Rightarrow \dfrac{\log 17}{\log 3+\log 5}\times \dfrac{\log 1-\log 5}{\log 3}$
We check the logarithm table and put logarithmic value in the above step to have,
\begin{align} & \Rightarrow \dfrac{1.23}{0.477+.699}\times \dfrac{0-0.699}{0.477} \\ & \Rightarrow \dfrac{1.23\times \left( -0.699 \right)}{1.176\times 0.477}=\dfrac{-0.859}{0.560}=-1.49 \\ \end{align}
So the product of logarithms is $-1.49$ approximately and the integers between which $-1.49$ are -1 and -2.

Note: We note that if we logarithm with base 10 is called common logarithm and we have changed the base to 10 because we know about common logarithm, We note that base cannot be 1 or negative. If we do not want to use logarithm table we can use logarithmic identities inequality $b > 1,x > y\Rightarrow {{\log }_{b}}x < {{\log }_{b}}y$.