Please help me solve 10=1880(120/sqrt(x))^-1.46

[Rickus vd Berg]

Hi there

I am currently stuck with an equation that I just can not solve.

10=1880(120/sqrt(x))^-1.46

What is x?

 

[Deleted member 4993]

Hi there

I am currently stuck with an equation that I just can not solve.

10=1880(120/sqrt(x))^-1.46

What is x?

Since you know you are stuck - you must have made some attempts.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[Johulus]

You just need to play around with numbers a little.

\(\displaystyle 10=1880(\dfrac{120}{\sqrt{x}})^{-1.46} \\ \dfrac{1}{188}=\dfrac{120^{-1.46}}{\sqrt{x^{-1.46}}} \\ \sqrt{x^{-1.46}}=188 \cdot 120^{-1.46} \\ x^{-1.46}=188^2 \cdot (120^2)^{-1.46}\)
Next step is to write \(\displaystyle 188^2 \) as \(\displaystyle (188^b)^{-1.46} \), where \(\displaystyle b \cdot (-1.46)=2 \). From this point on you should be able to finish it on your own.

 

[pka]

You just need to play around with numbers a little.
\(\displaystyle 10=1880(\dfrac{120}{\sqrt{x}})^{-1.46} \\ \dfrac{1}{188}=\dfrac{120^{-1.46}}{\sqrt{x^{-1.46}}} \\ \sqrt{x^{-1.46}}=188 \cdot 120^{-1.46} \\ x^{-1.46}=188^2 \cdot (120^2)^{-1.46}\)

Really?
\(\displaystyle 10=1880\left(\dfrac{120}{\sqrt{x}}\right)^{-1.46} \\ 1=188\dfrac{x^{0.72}}{(120)^{1.46}}\)

 

[Johulus]

Really?
\(\displaystyle 10=1880\left(\dfrac{120}{\sqrt{x}}\right)^{-1.46} \\ 1=188\dfrac{x^{0.72}}{(120)^{1.46}}\)

There are more ways to solve this task. It is irrelevant if you come at the correct result. This way it is not necessary to find the root by pluging it into a calculator. I don't see what you are trying to say.

\(\displaystyle x^{-1.46}=(188^{-\dfrac{100}{73}} \cdot 120^2)^{-1.46} \\ x=188^{-\dfrac{100}{73}} \cdot 120^2 \).

 

[Rickus vd Berg]

Hi guys. Thank you very much for al the positive feed back. This is actually an equation o work out peak particle velocity of ground vibration.

PPV=a(D/sqrt(E))^-b

a = ground constants
D = Distance
E = Total explosives per delay
b = slope value of ground constants

I already have my PPV value (10 mm/s) so I only had to figure out how many explosives I need per delay to reach a vibration level of 10mm/s

10=1880(120/sqrt(x))^-1.46

Thanks to global standards and MathPapa, I finally solved it.

sqrt(x) = 3.33

x = 11.08

I worked it backwards to get my answer. but I still need to show all my steps and that's my next problem. The way I got my answer is incorrect.

Initially, I got this far:

10=1880(120/sqrt(x))^-1.46

10/1880=1880/1880(120/sqrt(x))^-1.46(0.685)

and that's where I'm now. LOL.

 

[stapel]

10=1880(120/sqrt(x))^-1.46

Thanks to global standards and MathPapa, I finally solved it.

sqrt(x) = 3.33

x = 11.08

I worked it backwards to get my answer. but I still need to show all my steps and that's my next problem. The way I got my answer is incorrect.

Initially, I got this far:

10=1880(120/sqrt(x))^-1.46

10/1880=1880/1880(120/sqrt(x))^-1.46(0.685)

and that's where I'm now.

Where did (0.685) come from?

Try using what you've learned in algebra and arithmetic. What is the simplified form of 1880/1880? What is the simplified form of 10/1880 (in fractional, exact form)?

Then you have:

. . . . .\(\displaystyle \dfrac{1}{188}\, =\, \left(\dfrac{120}{\sqrt{\strut x\,}}\right)^{-1.46}\)

What power will "cancel" the -1.46 on the right-hand side? Raise both sides to this power. (Leave in exact form; don't use calculator approximations yet!) Then what power will "cancel" the square root? Raise both sides to this power. Then "cross-multiply" to get the "x" on top (on the left-hand side) and everything else on the right-hand side. Now plug into the calculator. You'll find that the correct answer is very close to your "solved it" value.

 

[Rickus vd Berg]

Stapel. To be honest. I never had maths in school. I hated it. And strangely enough, I love it now. but still at this point, I have no clue what is what regarding maths. I learn and try to teach myself as I go along in my job, obviously with a lot of help from the internet, books and tutors.