
How do you evaluate ${{\log }_{15}}1$? \[\]
Answer
574.2k+ views
Hint: We recall the definition of logarithm with base $b$ and argument $x$ as ${{b}^{y}}=x\Leftrightarrow {{\log }_{b}}x=y$. Here we are given ${{\log }_{15}}1$ argument is 1 and base is 15 . We assume ${{\log }_{15}}1=y$ and use the definition of logarithm to find $y$. \[\]
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 $x$ is called argument of the logarithm. The argument of the logarithm is always positive $\left( x>0 \right)$ and the base is also positive and never equal to 1 $\left( b>0,b\ne 1 \right)$. We know that when base and argument are equal we have;
\[{{\log }_{b}}b=1\]
We know the logarithmic identity involving power $m\ne 0$ where $m$ is real number as
\[m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}\]
We are asked to evaluate the value of ${{\log }_{15}}1$. We see that here argument of the logarithm is $x=1$ and the base is $x=15$. Let us assume ${{\log }_{15}}1=y$. So by definition of logarithm we have
\[{{15}^{y}}=1\]
We know that from exponentiation that for any non-zero $a$ we have ${{a}^{0}}=1$. We take $a=15$ to have ${{15}^{0}}=1$. So we have ;
\[{{15}^{y}}={{15}^{0}}\]
We know that if base is equal in an equation then their exponents will be equal. So we have;
\[\begin{align}
& \Rightarrow y=0 \\
& \Rightarrow {{\log }_{15}}1=0 \\
\end{align}\]
Note: We can alternatively solve using the identities of logarithm. We can write the logarithm as
\[{{\log }_{15}}1={{\log }_{15}}{{15}^{0}}\]
Now we can use the logarithmic identity involving power $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$ for $m=0,x=b=15$
\[{{\log }_{15}}1={{\log }_{15}}{{15}^{0}}=0\times {{\log }_{15}}15\]
Now we use the logarithmic identity when base and argument are equal ${{\log }_{b}}b=1$ for $b=15$to have
\[{{\log }_{15}}1={{\log }_{15}}{{15}^{0}}=0\times {{\log }_{15}}15=0\times 1=0\]
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 $x$ is called argument of the logarithm. The argument of the logarithm is always positive $\left( x>0 \right)$ and the base is also positive and never equal to 1 $\left( b>0,b\ne 1 \right)$. We know that when base and argument are equal we have;
\[{{\log }_{b}}b=1\]
We know the logarithmic identity involving power $m\ne 0$ where $m$ is real number as
\[m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}\]
We are asked to evaluate the value of ${{\log }_{15}}1$. We see that here argument of the logarithm is $x=1$ and the base is $x=15$. Let us assume ${{\log }_{15}}1=y$. So by definition of logarithm we have
\[{{15}^{y}}=1\]
We know that from exponentiation that for any non-zero $a$ we have ${{a}^{0}}=1$. We take $a=15$ to have ${{15}^{0}}=1$. So we have ;
\[{{15}^{y}}={{15}^{0}}\]
We know that if base is equal in an equation then their exponents will be equal. So we have;
\[\begin{align}
& \Rightarrow y=0 \\
& \Rightarrow {{\log }_{15}}1=0 \\
\end{align}\]
Note: We can alternatively solve using the identities of logarithm. We can write the logarithm as
\[{{\log }_{15}}1={{\log }_{15}}{{15}^{0}}\]
Now we can use the logarithmic identity involving power $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$ for $m=0,x=b=15$
\[{{\log }_{15}}1={{\log }_{15}}{{15}^{0}}=0\times {{\log }_{15}}15\]
Now we use the logarithmic identity when base and argument are equal ${{\log }_{b}}b=1$ for $b=15$to have
\[{{\log }_{15}}1={{\log }_{15}}{{15}^{0}}=0\times {{\log }_{15}}15=0\times 1=0\]
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