Technically part of my physics homework, but figured this would still apply here. I'm asked to verify the following equation (which I believe just means to do the algebra and get into this form). Note that $z \ll 1$

$$m-M\approx 43.23 - 5\log_{10} {\left(\frac{H_0} {68 \ \mathrm{kms^{-1}Mpc^{-1}}}\right)} + 5\log_{10}{z}+ 1.086(1-q_0)z $$

I am given the relation

$$m-M\approx 5\log_{10}\left( \frac{d_L}{1 \ \mathrm{Mpc}} \right)+ 25$$

As well as for small $z$

$$d_L \approx \frac{c}{H_0}z \left( 1 + \frac{1-q_0}{2}z \right)$$

I start with plugging in $d_L$

$$m-M \approx 5\log_{10} \left( \frac{1}{1 \ \mathrm{Mpc}} \cdot \frac{c}{H_0}z \left( 1 + \frac{1-q_0}{2}z \right) \right) + 25 $$

To make this easier to work with, let $\alpha=c/(H_0 \cdot1\ \mathrm{Mpc})$. I then split my logarithm

$$m-M\approx 5\log_{10}(\alpha)+5\log_{10}\left[z\left(1+\frac{1-q_0}{2}z\right)\right]+25$$

I analyzed the $z$ term individually using the relation $\log_{10}(1+x) \approx .4343x$ (given to me by my textbook)

$$5\log_{10}\left[z\left(1+ \frac{1-q_0}{2}z\right)\right]=5\log_{10}z+5(.4343\cdot\frac{1-q_0}{2}z) = 5\log_{10}z + 1.086(1-q_0)z$$

Now, I have

$$m-M\approx 5\log_{10}(\alpha)+25+5\log_{10}z + 1.086(1-q_0)z$$

Let's take a look at $5\log_{10}(\alpha)$. (Note $c=2.998\times10^8 \mathrm{ms^{-1}}$)

$$5\log_{10}(\alpha) = 5\log_{10}\left( \frac{c}{1 \ \mathrm{Mpc} \cdot H_0} \right) = 5\log_{10}c-5\log{10}{(1\ \mathrm{Mpc} \cdot H_0)} = 42.38 - 5\log_{10}{(1\ \mathrm{Mpc} \cdot H_0)}$$

Finally, I arrive to where I'm stuck.

$$m-M \approx 42.38 - 5\log_{10}{(1\ \mathrm{Mpc} \cdot H_0)}+25 + 5\log_{10}z + 1.086(1-q_0)z$$

I'm not sure how to get that last term that's in the equation I have to verify. The last bit of information I can offer is $H_0 = 68 \pm 2 \mathrm{kms^{-1}Mpc^{-1}}$, which confuses me since that would mean that logarithm in the equation should effectively be zero. If I don't have any luck here, I'll resort to Physics SE.



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