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I am meant to make $d$ the subject and simplify this as much as possible:

$$\ln(t) = -7.5 + 1.5\ln(d)$$

I originally had the answer as;

$$\ln\left(d\right)=\frac{\ln\left(t\right)+7.5}{1.5}$$

But using $t = 1.9$, I get $5.427$ something instead of $228$.

An online calculator, "WolframAlpha" shows the answer as $e^5t^{\frac{2}{3}}$

The $t^{\frac{2}{3}}$ seems to be the $1.5\ln(d)$ but I don't know how $1.5$ became a power of $t$.

The $e^5$ seems to be ${\frac{7.5}{2.5}}$ though I don't know where the $2.5$ came from.

an4s
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Anon
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3 Answers3

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$$ d=e^{\frac{\ln(t)+7.5}{1.5}}=e^{\frac{2}{3}(\ln(t)+7.5)}=e^{\frac{2}{3}\ln(t)} e^{\frac{2}{3}7.5}=t^\frac{2}{3} e^{15/3} $$

Note that $t^c$ is defined as $e^{c\log(t)}$.

cangrejo
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A few pointers:

  • $c\ln(a) = \ln\left(a^c\right)$
  • $c = c\ln(e)$
  • $\ln(a) + \ln(b) = \ln(a\cdot b)$

Combining the above,

$$\begin{align}\ln(t) &= -7.5 + 1.5\ln(d)\\ \ln(t) + 7.5\ln(e) &= 1.5\ln(d) \\ \frac{1}{1.5}\ln(t) + \frac{7.5}{1.5}\ln(e) &= \ln(d)\\ \frac23\ln(t)+5\ln(e) &=\ln(d)\\ \ln\left(t^{\frac23}\right)+\ln\left(e^5\right)&=\ln(d)\\\ln\left(e^5t^{\frac23}\right)&=\ln(d)\\\implies d&=e^5t^{\frac23}\end{align}$$

an4s
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$$\ln t=-7.5+1.5\ln d \\ 1.5\ln d=\ln t+7.5\\ \ln d=\dfrac{\ln t}{1.5}+5 \\ e^{\ln d}=e^\frac{\ln t}{1.5}e^5 \\ d=t^\frac23e^5$$

MafPrivate
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