Answer
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Hint: To solve this question, we will first change base of log as 'e' using the formula ${{\log }_{a}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}a}$ then we will convert both numerator and denominator obtained to exponential using the formula of exponential to log or vice versa given as ${{\log }_{a}}b=k\Rightarrow {{a}^{k}}=b$ so value of k is determined.
Complete step-by-step answer:
We are given that \[{{\log }_{10}}1=k\]
Observe that, here base of log is 10 and not 'e' so we will convert base of log into 'e' using the formula;
\[{{\log }_{a}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}a}\]
Comparing the above formula from expression ${{\log }_{10}}1$ we have a = 10 and b = 1
Applying the above stated formula we get:
\[\begin{align}
& {{\log }_{10}}1=k \\
& \dfrac{{{\log }_{e}}1}{{{\log }_{e}}10}=k\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
We have a formula for converting logarithmic function to exponential function, which is given as:
\[\begin{align}
& {{\log }_{a}}b=x \\
& {{a}^{x}}=b \\
\end{align}\]
So using this in LHS of equation (i) we get, let ${{\log }_{e}}1=t$
Using above formula:
\[{{e}^{t}}=1\Rightarrow t=0\text{ as }{{\text{e}}^{\text{0}}}\text{=1}\]
Again for denominator let ${{\log }_{e}}10=b$ then
\[{{e}^{s}}=10\]
s can be undetermined value (let it be s only). So, value of equation (i) is
\[\begin{align}
& \dfrac{{{\log }_{e}}1}{{{\log }_{e}}10}=\dfrac{t}{s}=k \\
& \dfrac{{{\log }_{e}}1}{{{\log }_{e}}10}=\dfrac{0}{s}=0 \\
& k=0 \\
\end{align}\]
Hence the value of k = 0.
Therefore, the value of $\text{5k}\Rightarrow 5\times 0=0$
Therefore, 0 is the answer.
Note: Another method to solve this question can be, not converting ${{\log }_{10}}1$ into log of base e and directly applying the formula of conversion of log into exponential. Doing so we get:
\[{{\log }_{10}}1=k\Rightarrow {{10}^{k}}=1\]
Now since any power of 10 cannot become apart from k being 0 as ${{10}^{0}}=1$
Value of k = 0 $\Rightarrow 5k=0$ which is the correct answer.
Complete step-by-step answer:
We are given that \[{{\log }_{10}}1=k\]
Observe that, here base of log is 10 and not 'e' so we will convert base of log into 'e' using the formula;
\[{{\log }_{a}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}a}\]
Comparing the above formula from expression ${{\log }_{10}}1$ we have a = 10 and b = 1
Applying the above stated formula we get:
\[\begin{align}
& {{\log }_{10}}1=k \\
& \dfrac{{{\log }_{e}}1}{{{\log }_{e}}10}=k\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
We have a formula for converting logarithmic function to exponential function, which is given as:
\[\begin{align}
& {{\log }_{a}}b=x \\
& {{a}^{x}}=b \\
\end{align}\]
So using this in LHS of equation (i) we get, let ${{\log }_{e}}1=t$
Using above formula:
\[{{e}^{t}}=1\Rightarrow t=0\text{ as }{{\text{e}}^{\text{0}}}\text{=1}\]
Again for denominator let ${{\log }_{e}}10=b$ then
\[{{e}^{s}}=10\]
s can be undetermined value (let it be s only). So, value of equation (i) is
\[\begin{align}
& \dfrac{{{\log }_{e}}1}{{{\log }_{e}}10}=\dfrac{t}{s}=k \\
& \dfrac{{{\log }_{e}}1}{{{\log }_{e}}10}=\dfrac{0}{s}=0 \\
& k=0 \\
\end{align}\]
Hence the value of k = 0.
Therefore, the value of $\text{5k}\Rightarrow 5\times 0=0$
Therefore, 0 is the answer.
Note: Another method to solve this question can be, not converting ${{\log }_{10}}1$ into log of base e and directly applying the formula of conversion of log into exponential. Doing so we get:
\[{{\log }_{10}}1=k\Rightarrow {{10}^{k}}=1\]
Now since any power of 10 cannot become apart from k being 0 as ${{10}^{0}}=1$
Value of k = 0 $\Rightarrow 5k=0$ which is the correct answer.
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