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I would like to use the calculation rules for exponential function $$1)\quad e^{x+y}=e^{x}\cdot e^{y}\quad\text{ and }\quad2)\quad(e^a)^b=e^{a\cdot b}$$ to derive the following calculation rules for the natural logarithm: $$\ln(x\cdot y)=\ln(x)+\ln(y)\quad\text{ and }\quad \ln(a^b)=b\cdot\ln(a)$$ My solution so far looks like this:

1)\begin{align*} e^{x+y}&=e^{x}\cdot e^{y}& \ln()\text{ on both sides}\\ \ln(e^{x+y})&=\ln(e^x\cdot e^y)\\ x+y&=\ln(e^x\cdot e^y) \end{align*}

From here, I don't know any further.

2)\begin{align*} (e^a)^b&=e^{a\cdot b}& \ln()\text{ on both sides}\\ \ln(e^a)^b&=\ln(e^{a\cdot b})\\ \ln(e^a)^b&=a\cdot b \end{align*}

I can't get any further here either. I really appreciate your help and tips. Thanks in advance!

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  • $\begingroup$ Use $x=e^{\ln x}$. Then $\ln(xy)=\ln(e^{\ln x}e^{\ln y})=...$ $\endgroup$ Jan 16 at 22:42
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    $\begingroup$ It should be $\ln\left(a^b\right)=b\ln(a)$ instead of $\ln(a)^b=b\ln(a)$. $\endgroup$ Jan 16 at 22:45

3 Answers 3

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If $x=e^a$ and $y=e^b$, then\begin{align}\ln(xy)&=\ln\left(e^ae^b\right)\\&=\ln\left(e^{a+b}\right)\\&=a+b\\&=\ln(x)+\ln(y).\end{align}And\begin{align}\ln\left(a^b\right)&=\ln\left(e^{b\ln a}\right)\\&=b\ln(a).\end{align}

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Let $s=\ln(a)$ and $t=\ln(a^b)$

Then $a=e^s$ and $a^b=e^t$.

But $a^b=(e^s)^b=e^{sb}$

So $e^t=e^{sb}$

Thus $t=sb$.

That is, $\ln (a^b)=b\ln(a)$.

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You have done it let me finish your equations:

\begin{equation} \begin{array}{l} e^{x+y}=e^{x} \cdot e^{y} \Leftrightarrow \ln \left(e^{x+y}\right)=\ln \left(e^{x} \cdot e^{y}\right) \\ \Leftrightarrow \ln(e^x \cdot e^y) = \ln(e^{(x+y)}) = x + y = \ln(e^x) + \ln (e^y). \end{array} \end{equation}

$$\left(e^{a}\right)^{b}=e^{a \cdot b} \Leftrightarrow \ln \left(\left(e^{a}\right)^{b}\right)=\ln \left(e^{a \cdot b}\right)=a \cdot b = b \cdot \ln(e^a).$$

I used the injectivity of the natural logarithm and that if $f(x) = e^x \Rightarrow f^{-1}(x) = \ln(x).$

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