Find d

[johans]

Hi,

My maths from 20 years back I have forgotten.

I need to find the value of d


FSPL = -93
f = 868000
GTX = 6
GRX =3
c = 299792458

I cant recall how to manipulate the formula to find d, please help

 

[topsquark]

Hi,

My maths from 20 years back I have forgotten.

I need to find the value of d

View attachment 32272
FSPL = -93
f = 868000
GTX = 6
GRX =3
c = 299792458

I cant recall how to manipulate the formula to find d, please help

You essentially have
$log_{10}(d) = A$
where A is some constant.

What is $10^{log_{10}(d)}$? (Hint: $log_{10}(x)$ is the inverse function of $10^x$.)

-Dan

 

[The Highlander]

You essentially have
$log_{10}(d) = A$
where A is some constant.

What is $10^{log_{10}(d)}$? (Hint: $log_{10}(x)$ is the inverse function of $10^x$.)

-Dan

Maybe his/her difficulty is in finding "A". ?

 

[Harry_the_cat]

Hi,

My maths from 20 years back I have forgotten.

I need to find the value of d

View attachment 32272
FSPL = -93
f = 868000
GTX = 6
GRX =3
c = 299792458

I cant recall how to manipulate the formula to find d, please help

Can you simplify

$\displaystyle -93 = 20 \log_{10}(d) + 20 \log_{10}(868 000) + 20\log_{10}(\frac{4\pi}{299 792 458}) - 6 - 3$

down to

$\displaystyle -93 = 20\log_{10}(d) + ??$

 

[JeffM]

The basic idea is this definition of logarithms.

$$\text {Given } \alpha > 0, \ \alpha \ne 1, \text { and } \beta > 0, \\ \log_{\alpha}(\beta) = \gamma \iff \beta = \alpha^{\gamma}.$$
(From that definition, topsquark’s hint follows directly, but I actually think it is easier to work from the definition itself.)

The Cat has given you one method, but I find it a bit easier to go

$$-93 = 20 \log_{10}(d) + 20 \log_{10}(868000) + 20 \log_{10} \left ( \dfrac{4 \pi}{299792948} \right ) - 3 - 6 \implies \\ - 84 = 20 \log_{10} + 20 \log_{10} \left ( \dfrac{3.472 * 10^6 * \pi}{2.997,929,480 * 10^8} \right ) \implies\\ \log{10}(d) + \log_{10} \left ( \dfrac{3.472 * \pi}{2.997,929,480} * 10^{-2} \right ) = -\dfrac{84}{20} \implies\\ \log_{10}(d) - 2 \log_{10} \left ( \dfrac{3.472 * \pi}{2.997,929,480} \right ) = - 4.2.$$
Now can you finish it up?

 

[Harry_the_cat]

The basic idea is this definition of logarithms.

$$\text {Given } \alpha > 0, \ \alpha \ne 1, \text { and } \beta > 0, \\ \log_{\alpha}(\beta) = \gamma \iff \beta = \alpha^{\gamma}.$$
(From that definition, topsquark’s hint follows directly, but I actually think it is easier to work from the definition itself.)

The Cat has given you one method, but I find it a bit easier to go

$$-93 = 20 \log_{10}(d) + 20 \log_{10}(868000) + 20 \log_{10} \left ( \dfrac{4 \pi}{299792948} \right ) - 3 - 6 \implies \\ - 84 = 20 \log_{10} + 20 \log_{10} \left ( \dfrac{3.472 * 10^6 * \pi}{2.997,929,480 * 10^8} \right ) \implies\\ \log{10}(d) + \log_{10} \left ( \dfrac{3.472 * \pi}{2.997,929,480} * 10^{-2} \right ) = -\dfrac{84}{20} \implies\\ \log_{10}(d) - 2 \log_{10} \left ( \dfrac{3.472 * \pi}{2.997,929,480} \right ) = - 4.2.$$
Now can you finish it up?

Your second line is missing a d.
Your third line need 10 to be subscript.
This is a bit confusing for someone who has pretty much admitted they've forgotten about logs.

 

[johans]

Thank you all, jolted the old memory banks.

 

[JeffM]

Thanks Harry. I managed to pack a typo into each line. That may be a record even for me.

Second line should read

$$- 84 = 20 \log_{10}(d) + 20 \log_{10} \left ( \dfrac{3.472 * 10^6 * \pi}{2.997,929,480 * 10^8} \right ) \implies$$
Third line should read

$$\log_{10}(d) + \log_{10} \left ( \dfrac{3.472 * \pi}{2.997,929,480} * 10^{-2} \right ) = - \dfrac{84}{20} \implies$$
The fourth line should read

$$\log_{10}(d) - 2 + \log_{10} \left ( \dfrac{3.472 * \pi}{2.997,929,480} \right ) = - 4.2.$$
I apologize johans.