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I was messing around with my calculator and meant to put in $\log(9)/\log(3)$, which was supposed to be two. I look back, and I see my mistake, but curiously, I input that to the function $e^{(x)}$ and it comes out with $100$... Why is that? Why is $e^{\ln(9)/\log(3)}=100$?

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    $\begingroup$ We usually put a backslash before log, just as you did for ln, to render it in the function font. $\endgroup$ Commented Dec 21, 2013 at 21:30

7 Answers 7

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In the calculator $\log$ is for the decimal logarithm and $\ln$ for the natural logarithm so

$$\frac{\ln 9}{\log3}=\frac{2\ln 3\times \ln 10}{\ln3}=\ln 100$$

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    $\begingroup$ How many hats are here! Morning Sami. ;+) $\endgroup$
    – Mikasa
    Commented Dec 22, 2013 at 3:11
  • $\begingroup$ You nailed it! +1 $\endgroup$
    – amWhy
    Commented Dec 22, 2013 at 15:58
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We can convert $\log(3)$ to base $e$ with the change of base rule: $$\log(3) = \frac{\ln(3)}{\ln(10)}$$ Then the expression $(\ln(9)/\log(3))$ equals $\frac{ln(9)}{\ln(3)/\ln(10)}$, which is equal to $$\begin{align} \ln(9) \cdot \frac{\ln(10)}{\ln(3)} &= 2\ln(3) \cdot \frac{\ln(10)}{\ln(3)}\\ &= 2 \cdot \ln(10)\\ &= \ln(100) \end{align}$$

So $e^{\ln(9)/\log(3)} = e^{\ln(100)}$. We know $e^{\ln(a)} = a$, so $e^{\ln(100)} = 100$.

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  • $\begingroup$ So, would this function $y=abs(e^{ln(x^2)/log(x)})$ model it? $\endgroup$
    – user16795
    Commented Dec 21, 2013 at 22:42
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Hint: $$\log(3)=\log_{10}(3)=\frac{\ln(3)}{\ln(10)}\,.$$

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To take it one step at a time, recall the definition of the logarithm.

In particular, $\ln9$ is the (unique) number that satisfies $\mathrm{e}^{\ln9}=9$. Also, $\log3$ is defined by $10^{\log3}=3$.

In this way, $$\mathrm{e}^{\ln9/\log3}=\left( \mathrm{e}^{\ln9}\right) ^{1/\log 3}=9^{1/\log3}$$ and $$9^{1/\log3}=100$$ because $$100^{\log3}=10^{2\log3}=\left( 10^{\log3}\right) ^{2}=3^{2}=9.$$

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Because $\ln(9)=\log(9)\ln(10)=2\log(3)\ln(10)$, so $\ln(9)/\log(3)=2\ln(10)$

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  • $\begingroup$ Why is $\ln(9)=\log(9)\ln(10)$ ? $\endgroup$
    – user16795
    Commented Dec 21, 2013 at 21:36
  • $\begingroup$ @user16795: $e^{\ln(9)}=9$ but also $e^{\log(9)\ln(10)}=\left(e^{\ln(10)}\right)^{\log(9)}=10^{\log(9)}=9$. Since $e^x$ is injective, it follows that the exponents are equal. $\endgroup$ Commented Dec 21, 2013 at 21:37
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$\log(e^{\ln((9)/\log(3))})=\dfrac{1}{\log3}\log(e^{\ln9})=\dfrac{1}{\log3}\log9= 2\dfrac{\log3}{\log3}=2$

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So this is the equation: $y=abs(e^{(ln(x^2)/log(x)})$

@Newb I'm sorry for taking your work, I'm not familiar with this.

We can convert log(x) to base e with the change of base rule: $\log(x) = \frac{\ln(x)}{\ln(10)}$

Then the expression $(\ln(x^2)/\log(x)) equals \frac{ln(x^2)}{\ln(x)/\ln(10)}$, which is equal to \begin{align} \ln(x^2) \cdot \frac{\ln(10)}{\ln(x)} &= x\ln(x) \cdot \frac{\ln(10)}{\ln(x)}\\ &= x \cdot \ln(10)\\ &= x\ln(10) \end{align} So $e^{\ln(x^2)/\log(x)} = e^{x\ln(10)}$. But when you graph this equation, it doesn't equal 100...? Am I doing something wrong?

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