Question

How do you evaluate ${{\log }_{15}}1$? $$$$

Answer 1

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;

& \Rightarrow y=0 \\\ & \Rightarrow {{\log }_{15}}1=0 \\\ \end{aligned}$$ **Note:** We can alternatively solve using the identities of logarithm. We can write the logarithm as $${{\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$$