# Analytical question on year calculation.

I don't know, how to solve the question. I request an expert to help me on solving.

Climbing vine can grow double in height every year. Since planting the vine 6 years ago, it has grown 25 feet. How many years did it take to grow to half this height ?

To the people above me: for all intents and purposes your answers are close enough saying $6-1$ years, but it is not mathematically valid as the OQ states by how much the vine's height has increased, not what the current height is at $n=6$.

This question is a classic example of what is called exponential growth.

The vine starts off with some finite height $h_0$ when you plant it, and then on this day every year the vine is now at twice its original height. So the height at any year $n$ is given by $$h_n=2\cdot h_{n-1}$$

Through some recursion you notice that there is a general expression for the height at any time, based on the original height. This should make sense as the vine is doubling from its original height:

$$h_n=h_0\cdot 2^n$$

The information you are given is that the vine has increased in height by 25 ft by the time it is six years old. So, the current height is $$h_6=h_0+25$$. To make our math prettier let's just assign a parameter $\Delta h$ to be the change in height. So, $$\Delta h=25$$. Working from the second equation: $$h_6=h_o\cdot 2^n$$ $$\left(h_0+\Delta h\right)=h_0\cdot 2^n$$

We can solve for the initial height as follows: $$\Delta h=h_0\left(2^6-1\right) \implies h_0 = \frac{\Delta h}{2^6-1}$$

Now that we know the initial height of the vine, we can determine how long it took for it to reach half of the current height $h_n=\frac{\Delta h+h_0}{2}$

$$h_n=\frac{\Delta h +h_0}{2}=h_0\cdot 2^n$$

Now the question is a trivial example of substituting in $h_0$ and using the laws of logarithms to find the value of n. This would be a good practice for you and I'll let you know the end result is $$n=\frac{11 \log 2-2 \log 3-\log 7}{\log 2}$$

Approximately 5.02272 years. So, as I said 5 years is close enough but not correct.

• Approximate is more than enough. But i didnt expect that calculation was little complicated. Thank you for taking your time and explaining. I dont know all these expressions. Mar 29 '14 at 22:11

The height doubles every year, so the plant is at half of its current height in the previous year.

If we let $h_n$ denote the height of the vine after $n$ years, you see that $$h_{n+1} = 2 h_n,$$ which implies $$h_n = c \cdot 2^n, \quad \text{for } n \geq 0.$$ Notice that $$h_0 = c \cdot 2^0 = c,$$ so $$h_n = h_0 \cdot 2^n, \quad \text{for } n \geq 0.$$

Can you take it from here?

• Do you have the answer for that question. I am not sure, whether your steps would be helpful Mar 27 '14 at 20:58
• @Medex I do. I also found an easier way to solve it. The hint is, it doubles in value each year and you need to find how long it took to grow to half the height at 6 years. Mar 27 '14 at 22:24
• answer is 5 years. But i dont know, How to do it. I have the question and answer. Mar 28 '14 at 1:23
• @Medex well, it grows to 25 in 6 and tasks each year to double. So to grow half the height must have taken one year because it was half the height and then doubled in one year to reach the full height, so half the height was at 6-1=5 years Mar 28 '14 at 2:42
• It grows by 25ft in six years, not to a height of 25ft. Mar 29 '14 at 21:41