Exponential performance decreases with load

Question

In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is?

Using the standard equation f(x) = a * b^x, I figured that when a = 1000000 and b = 8912509381337455, the answer would be 46.02.

Is this correct a fine approach? However it involves using calculator.

Further, this was a multiple choice question in some exam, and the options given were:

a) 40.00 b) 46.02 c) 60.01 d) 92.02.

I saw an explanation using the process of elimination as follows:

(down) load cycles for failure (increase) exponentially

Load = x
Cycles for failure = y
x/2 = y^2, x/3 = y^3, x/4 = y^4

Hence,

Option a) 80/2 = 100 ^ 2 = 10,000 cycles
Option d) Load is >80 hence it is not possible
Option c)
Load = 60.01 = 3/4 (80)
From the given relation,
3/4 (80) = 10 ^ 4/3 = 464 cycles which is not right

So the answer is option b)

Is this right?

I mean is the relation x/2 = y^2, x/3 = y^3, x/4 = y^4 correct?

I would have accepted if it is x/2 = y^1/2, X/3 = y^1/2, x/4 = y^1/4.

Answer 1

The derivation is right $\frac{3x}{4} = \frac{\frac{3x}{2}}{2} = y^{\frac{4}{3}}$. Hence it is valid.

The general relationship is as below:

x = number of loads

y = cycles of failure

The realtionship is

$\frac{x}{b} = y^b$

according to the first and third statement. The first statement is the opposite way of expressing the explanation given in the third statement if you notice. That is why you are confused.

Edit:

The correct way:

$100 = N_0 e^{-80\lambda}\tag{1}$

$10000= N_0 e^{-40\lambda}\tag{2}$

From the above we get, $\lambda = \frac{ln(100)}{40}=0.11515$ $

$$N_0 = 100*e^{2ln(100)} = 1000000$$

$$N(t) = 1000000*e^{\left(-46.02*0.11515\right)} = 5000$$

Thus this is the right model for this

The earlier one is done to skin the cat without working out the details.

Answer 2

What you did is good. Set up f(x) = a * b^x.

If no calculator is available then one can exclude a) c) and d) using simple estimations.

But you are right. This is crazy: x/2 = y^2, x/3 = y^3, x/4 = y^4.

Full calculation for reference.

h(80) = 100
h(40) = 10000

h = 10^(A*x+B)

2 = 80A + B
4 = 40A + B

A = -2/40 = -1/20
B = 6

h = 10^(6-x/20)

h = 5000

lg(5000) = 6-x/20
x = 20*(6-lg(5000)) = 20*(3-lg(5)) = 46.020599913279625