Solving for L when L^.68 = S^(-.16)?

Question

Im trying to solve for L in terms of S but I can't seem to figure out the algebra needed to solve for L when L ^.68 = S^(-.16). Anyone know how to solve for L?

Answer 1

taking logon bothsides
0.68 log L=-0.16 logS
log L=(-0.16/0.68)logS

Answer 2

I will make this very simple for you:

L^0.68 = L^(68/100) = S^(-0.16) = S^(-16/100)

Now put both sides of the equation to the power 100/68:


[L^(68/100)]^(100/68) = [S^(-16/100)]^(100/68)

You will get: L = S^(-16/68) = S^(-4/17).

You can double-check:

For S=67:

L= S^(-4/17) = 0.371822.

L^0.68 = 0.371822^0.68 = 0.510303 and

S^-0.16 = 67^-0.16 = 0.510303.

Both numbers match thank God!!!

Answer 3

L^ (68/100)=S^(-16/100)
Now Raise both sides to the power 100/68
Hence L=[S^(-16/100)]^(100/68)
i.e. L= S^ (-16/68)
i.e. L= S^(-4/17)
i.e L=1/[s^(4/17)]
what else do you want??????????

Answer 4

x^2 - 10x = -sixteen upload sixteen to the two facet x^2 - 10x + sixteen = 0 then locate 2 numbers that multiply to offer sixteen and upload to offer -10. -8 and -2 artwork as -8x-2=sixteen and -8 + -2 = -10 so x^2 - 8x - 2x + sixteen = 0 (x-8)(x-2)=0 desire this permits you!

Answer 5

L^(.68) = S^(-.16)

L^(68/100) = S^(-16/100)

raise both sides by 100

L^(68) = S^(-16)

68thrt both sides

L = S^(-16/68)

L = S^(-4/17)
or
L = 1/(S^(4/17))

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Another way you can write this is

L^(.68) = S^(-.16)

.68ln(L) = -.16ln(S)
ln(L) = (-.16/68)ln(S)
ln(L) = (-4/17)ln(S)
L = e^((-4/17)ln(S))
L = (e^(ln(S)))^(-4/17)

the e and ln cancel out

L = S^(-4/17) or 1/(S^(4/17))

Answer 6

L^.68 = S^(-.16)
Raise the power both side by (1/0.68) both sides
{L^0.68}^(1/0.68) = {S^(-0.16)}^(1/0.68)
L =S^{-0.16/0.68}
= S^(-4/17)

Answer 7

use logarithms as the poster above said and then exponentiate above e (euler's constant)
so
Ln(L) = (-.16/0.68)Ln(S)
e^Ln(L) = e^(-.16/0.68)Ln(S)
e = 2.71.......

e^Ln(x) = x, so:

L = e^(-.16/0.68)Ln(S)

Answer 8

perfect.