Question

Proof it
$\boxed{\rm log_n1=0}$​

Answer 1

$\huge{\rm log_n1=0}$

Proof :-

$\huge{\rm log_n1=0}$

Since, ${a}^{0}$ = 1

It's inverse,

0 = Log 1 n

n > 0 ,

➝ $\dfrac{ {n}^{1} }{ {n}^{1} }$

➝ ${n}^{1 - 1}$

➝ ${n}^{0} = 0$

Hence, $\huge{\rm log_n1=0}$

We know,

Log 1 = 0

It means that the logarithm of 1 is always 0. It doesn't depend on the base of the logarithm. No matter it may be a,b,c,d..

It's because any number to the power 0 is 0.

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$\boxed{\boxed{\begin{minipage}{4.2cm}\circ\sf\:\ln 1=0\\\circ\sf\:\ln e=1\\\circ\sf\:\ln x=y\leftrightarrow e^y=x\\\circ\sf\:e^{\ln x}=x,\:x>0\\\circ\sf\:\ln(e^x)=x,\:x\in\mathbb{\large R}\\\circ\sf\:\ln(xy)=\ln x+\ln y\\\circ\sf\:\ln(x/y)=\ln x-\ln y\\\circ\sf\:\ln(x^r)=r\ln x\\\circ\sf\:\ln x=log_e\:x\end{minipage}}}$