ExponentsAndLogs Learning Objectives

# Problem 1 of 3

(a) Solve the equation 3^{x+1} = *e*(5^{x}) algebraically.

Express your final answer correct to 3 decimal places. You may use a calculator to find the final answer.

(b) Solve algebraically: 2log_{3}(*x* + 4) - log_{3}(-*x*) = 2

Solve this equation **without** using a calculator.

**Solution**

A) 3^{(x+1)}=e(5x)

-put the log in front of the base the three

log3^{(x+1)}=loge(5x)

-put the exponent on front of the base

(x+1)log3=loge(5x)

-doing that you are now able to multiply the log3 with the brackets since the brackets are like numbers

xloge3+log3=loge(5x)

-put all the X's to one side

xlog3+5x=log3 <- (xlog3 - loge5x = log3)

-take out all the X's

x(log3+5)=log3

-get the X by itself and solve

x=log3/(log3+5)

=.1801

**Great job but I feel that there is an easier alternate solution to this problem. And I think your answer is incorrect.**

3^{(x+1)} = e(5x)

**ln both sides because there is e**

ln3^{(x+1)} = lne(5x)

**using the power of a power law the (x+1) comes down and
**

**ln of e = 1**

(x+1)ln3 = 5x

**distribute (x+1)**

xln3 + ln3 = 5x

**put all the x variable on one side**

ln3 = 5x - xln3

**factor out the x**

ln3 = x(5 - ln3)

**isolate the x**

x = ln3 / (5 - ln3)

**solve with calculator**

x = 0.282

B) 2log_{3}(x+4)-log_{3}(-x)=2

-use the quotient law

log_{3}((x+4)2)/-x=2

-then simplfy it use log

x^{2}+8x+16/-x=2

-the simplify cross multiply the with the -x and bring the 2 over

x^{2}+7x+14

I have a feeling its wrong please someone correct it

B)2log_{3}(x+4)-log_{3}(-x)=2

-use the quotient law

log_{3}((x+4)^{2})/-x=2

-Expand the binomial

log_{3}(x^{2}+8x+16)/-x=2

-Since 3 is the base, you would take 2 as the exponent

x^{2}+8x+16/-x=3^{2}

x^{2}+8x+16/-x=9

-cancel the denominator, by multiplying -x by both sides

x^{2}+8x+16=-9x

-bring the -9x over to the other side

0=x^{2}+17x+16

0=(x+16)(x+1)

- X is rejected for both of them because in the original equation, 2log_{3}(x+4) - log_{3}(-x) = 2, it is negative

Good try both of the previous problem solvers. I think i got the solution.

2log_{3}(x+4)- log_{3}(-x)= 2

expand the equation:

2log_{3}x + 2log_{3}4 - log_{3}(-x)= 2

move the x terms to one side:

2log_{3}x - log_{3}(-x) = 2 - log_{3}4

factor out x:

x(2log_{3}1 - log_{3}(-1)) = 2 - log_{3}

By looking at the equation now, you can notice that there is a log_{3}(-1). We know that there is no such solution. Since we know that the expression is false, there is no way of finding x. Therefor **x is undefined!
**

** **

**
**# Problem 2 of 3

Many people mistakenly believe that carbon dating can be used to determine the age of dinosaur bones. In fact, carbon-14 has a half life of about 5730 years which only makes it useful for dating events within the last 35 000 to 50 000 years. However, this half-life does make it useful to archeologists.

If the mammoth bones discovered in this article are found to have only 10% of the amount of uranium-233 (half-life 162 000 years) a live mammoth should have, how long ago did this mammoth die?

**Solution**

# Problem 3 of 3

At the begining of an experiment a colony of tribbles has 2100 members. After 4 hours the population has increased by 20%. At that time a biologist, Dr. Mc Coy, treats the colony with an agent that slows the colony's rate of growth. At the new growth rate, the colony takes three times as long to double in size as at the old growth rate. How long after the treatment will the size of the colony of tribbles reach 12 600?

**Solution**

A=12600 (the amount at the end of the time period t)

Ao=2100 (the original amount)

m=3 (growth rate)

t=4 (time that has passed)

p=?

A=Ao(m)^(t/p) (i used this formula because we r given lots of information)

12600=2100(3)^(4/p) (i just replaced in the numbers in the formula)

12600=6300^(4/p) (multiply 3 and 2100)

ln12600=ln6300^(4/p)(take (ln) of both sides)

ln12600=(4/p)ln6300 (bring the exponent in front)

pln12600=4(ln6300) (multiply both side by p)

p=(4(ln6300))/(ln12600) (divide so that p can be on its own)

p=3.7063 (use your calculator)

therefore the size of the colony of tribbles will reach 12 600 after 3.7063 hours.

## Comments (2)

## Anonymous said

at 11:28 pm on Jun 4, 2006

I don't know what happened to the coment that i leave here a few seconds ago but I want to make a comment for the one who did problem 3 of 3, i cannot write everything i wrote before but what i want to say is that the question is asking for the time between after the treatment and the colony to reach 12600. i got 105.93 hours as an answer for that.

## Anonymous said

at 11:31 pm on Jun 4, 2006

i dont know what happened to the comment i leave before but i hope this is enough for the constructive modification part cause its hard for me to explain again this stuff. sorry for not asking your permission cause its almost a few munites before the deadline.

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