How to make γ the subject of the formula? (logarithms involved)

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In summary, the conversation discusses the process of making γ the subject of a formula involving logarithms. The equation includes two instances of γ, one on each side of the equation within the logarithm. The variables involved are explained and the attempt at a solution is shown, but it is mentioned that there is no closed form for the equation and approximation methods will need to be used.
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Kelles
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How to make γ the subject of the formula? (logarithms involved)

Homework Statement



My problem here is that i want to make γ the subject of the formula so that i can calculate it since all other variables are know. As you can see from the equation below there are two γ inside the equation one on the left hand side and one on the right hand side of the equation which is inside the logarithm [log((qnet+γ)/pa)].

The variables involved are shown below, however you do not really need to know what each one means since the problem here is about mathematics and how to make γ subject of the formula if that's possible but i did include all variables to make it easier to understand the equation

γ = unit weight of soil = unknown
γW = unit weight of water = 9.81
Rf = friction ratio 0.5
qnet = net cone resistance = 5.53
pa = atmospheric pressure = 0.1

Homework Equations



Equation:γ/γW = 0.27 [log Rf] + 0.36 [log((qnet+γ)/pa)] +1.236

The Attempt at a Solution



γ/γW = 0.27 [log Rf] + 0.36 [log((qnet+γ)/pa)] +1.236

Substituting Rf = friction ratio 0.5

γ/γW = 0.27 [log 0.5] + 0.36 [log((qnet+γ)/pa)] +1.236

γ/γW = -0.08 + 0.36 [log((qnet+γ)/pa)] +1.236

γ/γW = -0.08 +1.236 + 0.36 [log((qnet+γ)/pa)]

γ/γW = 1.15 + 0.36 [log((qnet+γ)/pa)]

γ/γW - 1.15 = + 0.36 [log((qnet+γ)/pa)]

(γ/γW - 1.15)/0.36 = log((qnet+γ)/pa)

10(γ/γW - 1.15)/0.36 = (qnet+γ)/pa

I am not sure what to do next, i don't know how to get both γ together. Some help would be really appreciated.

Thank you.
 
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  • #2


There is no closed form for mixes of functions like this. You will have to use approximation methods, iterative usually.
 
  • #3


Ok thanks a lot for you reply
 

1. How do I make γ the subject of the formula when logarithms are involved?

To make γ the subject of the formula when logarithms are involved, you will need to use logarithmic rules and properties to isolate γ on one side of the equation. This will involve using the inverse of the logarithm, exponentiation, and algebraic manipulation.

2. Can you provide an example of making γ the subject of the formula with logarithms?

Sure, let's use the formula y = a*log(x) + b, where we want to make γ the subject. First, we can use the inverse of logarithms to get rid of the logarithm on the right side by taking the exponent of both sides. This gives us e^y = e^(a*log(x) + b). Then, we can use the property log(a*b) = log(a) + log(b) to expand the right side to e^y = x^a * e^b. Finally, we can divide both sides by e^b to get γ on its own: γ = (e^y)/x^a.

3. What is the importance of making γ the subject of the formula?

Making γ the subject of the formula allows us to solve for this variable and manipulate the equation to find its relationship with other variables. This can help us understand the behavior and properties of γ in the context of the equation. It is also important for simplification and making calculations easier.

4. Are there any tips for making γ the subject of the formula with logarithms?

One tip is to carefully apply logarithmic rules and properties, making sure to perform the same operation on both sides of the equation. Another tip is to use algebraic manipulation to isolate γ on one side of the equation. It may also be helpful to review logarithmic identities and practice solving equations involving logarithms.

5. How do I know if I have correctly made γ the subject of the formula?

You can check if you have correctly made γ the subject of the formula by substituting your solution back into the original equation and simplifying. The resulting equation should be equivalent to the original equation. You can also check your solution by graphing the original equation and your solution and seeing if they have the same shape and intersect at the same point.

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