# The length, L(t) of a snake after t months is given by `L(t)=Lo2^(0.5*t)` , where Lo is the initial length. How long will it take, to the nearest month, for the snake to reach six times its...

The length, *L(t)* of a snake after *t* months is given by `L(t)=Lo2^(0.5*t)` , where *Lo* is the initial length. How long will it take, to the nearest month, for the snake to reach six times its length?

A. 3 months

B. 5 months

C. 6 months

D. 2 months

Please Show me how to do it It's all a wanna know. The answer is B by the way!

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To solve any problem, it is always a good practice to list what is given and what is asked for in a particular problem.

1. What are given:

1a. The time-dependent (in months) equation that describes the growth of the snake: L(t) = L(0)*2^{0.5*t};

1b. Length reached after the required time (t = T) in months as compared to the initial length: L(T) = 6*L(0).

2. What is asked for: the time (T) in months so that L(T) = 6*L(0).

Just plug in the data into the given equation:

L(T) = L(0)*2^{0.5*T} -> 6*L(0) = L(0)*2^{0.5T} -> 6=2^{0.5*T}.

Now take ln on both sides:

ln 6 = ln{2^{0.5*T}} -> ln 6 = 0.5*T*ln2 from ln M^a = a*ln M

Isolate what is required on the left:

T = 2*{ln 6}:{ln2}

You can now use ln directly or convert it into log base ten to solve it. Using directly ln 6 = 1.79... and ln 2 = 0.69...

T = {1.79...}:{0.69...} = 5.19...

To the nearest integer this is 5 months. Alternative B).

L(t) = Lo * 2^(0.5*t)

To make the snake 6 times it's length, you multiple Lo (the initial length) by 6.

L(t) = Lo*6

Lo*6=Lo*2^(0.5*t) Divide Lo from both sides and they will cancel

6=2^(0.5*t) Log will get rid of your base 2

log(6)=log(2^(0.5*t))

log6=log2*0.5*t Use your calculator to evaluate log 6 and log 2

0.78=0.3*0.5*t Combine like terms and divide out to get t by itself

t= 5.2

So, the answer is B, 5 months.

Lo = initial length

L(t) after six months = L(6) = 6 * Lo

Given: L(t) = Lo * 2^0.5*t

So, plugging in the value of L(6) in the given equation, we get:

L(6) = Lo * 2^0.5*t

or 6 * Lo = Lo * 2^0.5*t

or 2^0.5*t = 6*Lo / Lo = 6

or log2^0.5*t = log 6 taking log on both sides

or 0.5*t* log2 = log6 as log(m)^n =n log(m)

or 0.5*t*0.301 = 0.778

or t = (0.778) / (0.5*0.301) = 0.778 / 0.1505 = 5.1694 ...

So, required time = **5 months** (answer B)

In order to understand this problem better, it is possible to simply substitute values for Lo and L(t) so that L(t) = 6Lo. In order to make it even simpler, it is best to use Lo = 1 and L(t) = 6.

With this, we get:

6 = 2^(0.5t)

Then, we can take the log of both sides in order to eliminate the exponent.

log6 = log2^(0.5t)

Using the power property of logs, the exponent of the 2 can be moved to the front, thus making the equation:

log6 = (0.5t)(log2)

Divide both sides by log2:

log6/log2 = 0.5t

Divide both sides by 0.5:

(log6/log2)/0.5 = t

t = 5.17 months

Fist, since we know that we want the final length to be 6 times the original, we set the equation equal to 6Lo.

6Lo = Lo 2^.5t

Divide by Lo, and solve for t, using the rules of logs:

6 = 2^.5t

log 6 = .5t * log 2

log 6/ log 2 * .5 = t

t = 5.1699

So 5 months