Question

13.5
Consider the logarithm function \(log=exp^{-1}\) on \(\mathbb{R}\). Prove:
c) \(log(xy) = log(x) + log(y)\)

Answer 1

\[\huge REMEMBER:\]
\[ \huge When,\\ \huge \quad { e }^{ y }=x \]
\[\huge In\quad base\quad e\quad LOGARITHM\quad of\quad 'x'\quad is: \]
\[\huge In(x)=\log_{ e }(x)=y \]
\[\huge e\quad \approx \quad 2.71828183 \]
\[\huge Ln\quad as\quad inverse\quad function\quad ofexponential\\ \huge \quad function:\]
\[\huge The\quad \mathbb{N}\quad logarithm\quad function\\ \huge \quad Ln(x)\quad is\quad the\quad inverse\quad function\\ \huge \quad of\quad the\quad exponential\\ \huge \quad function\quad ex.For\quad x>0,\]

Answer 2

@Master.RohanChakraborty thank you

Answer 3

\[
x= e^{log(x)}\\
y= e^{log(y)}\\

e^{log(xy)} =x y =e^{log(x)}e^{log(y)}= e^{log(x) +log(y}\\
log(xy) =log(x) + log(y)
\]

Answer 4

thank you mr Elias

Answer 5

yw

Answer 6

Elais is SIR Elias Saab

Answer 7

@Master.RohanChakraborty ok thanks for correction :)