Expressing $b^x$ as $e^{x \ln b}$ Can anyone explain the following equality please? 
$$b^x = e^{x \ln (b)}$$
I've verified it with values but I can't think of what the proof for this would be.
 A: Following @dxiv's comment, we have the following:


*

*By definition of logarithm,
$$b=e^{\ln b}$$

*So, 
$$b^x = (e^{\ln b})^x$$

*Using that, for any real numbers $a,b,c$, we have
$$(a^b)^c=a^{b\cdot c}$$

*We arrive to
$$b^x\ =\ (e^{\ln b})^x\ =\ e^{x\cdot\ln b}\,.$$



Also note, that - as @Jack D'Aurizio commented - usually this formula is the definition of the exponential $b^x$ for general real numbers (for rational exponent $x$, we can extend the familiar definition, but for real exponents we need to use either limits or this formula). 
A: $$b^x=e^{\ln(b^x)}=e^{x \ln b}$$
Having used the properties $x =e^{\ln x}$ (exponential and natural 
logarithm are the inverse function of each other) and the property of the logarithm $\ln(a^b)=b \ln a$.
A: $$b^x = e^{x \ln (b)}$$ take $\log_{b}{} $ both sides
$$\large \log_{b}{b^x} = \log_{b}{e^{x \ln (b)}}\\$$
$$x\large \log_{b}{b} = x \ln (b).\log_{b}{e}\\ \to
x\large \log_{b}{b} = x \frac{1}{\log_{b}{e}}.\log_{b}{e}\\
x \times 1=x \times \frac{\log_{b}{e}}{\log_{b}{e} \\x=x } \checkmark$$
