PreCalculus: solving log eqns, exponential eqns, etc

[sweetliljenny]

I need help with about 6 questions (desparately)
if anyone could help me out with them ASAP please reply now and then
i'll put up the questions

thank you =)

 

[pka]

We not mind readers. You must post the questions.

 

[sweetliljenny]

1)10 ^ log(2x+1) = ln2 (solving it exactly for x)

2) ln(x-2) + ln(2x-3) = 2 lnx

3) You have 5,000,000 bacteria and 45% of the bacteria are dying every minute, how long will it take to have less than 1,000 bacteria? (You are assuming continuous decay)

4) The half-life of carbon-14 is eight days.
a) determine the continuous decay rate
b) If a lab acquires 2 g of carbon-14, how much of this sample will remain after 20 days?
c) How long will it be until only 0.01 g remains?



thank you =)

 

[sweetliljenny]

**** sorry Question 2 is also be solved exactly for x

 

[pka]

\(\displaystyle \L
\begin{array}{l}
10^{\log (2x + 1)} = 2x + 1 \\
2x + 1 = \ln (2)\quad \Rightarrow \quad x = ? \\
\quad \\
\quad \\
\ln (x - 2) - \ln (2x - 3) = \ln \left( {\frac{{x - 2}}{{2x - 3}}} \right) \\
\ln \left( {\frac{{x - 2}}{{2x - 3}}} \right) = \ln (x^2 )\quad \Rightarrow \quad \frac{{x - 2}}{{2x - 3}} = x^2 \\
\end{array}\)

 

[sweetliljenny]

im sorry but for number 2 it was ln(x-2) + ln(2x-3) = 2lnx

and log(A x B) = log A + log B

which is why i have ln [ (x-2)(2x-3) ] = lnx ^ 2

so i factored (x-2)(2x-3) .. and got 2x ^2 - 3x - 4x + 6 = x ^ 2

which then is 2x ^ 2 - 7x + 6 = x ^ 2

i need help from thereafter though

 

[galactus]

3) You have 5,000,000 bacteria and 45% of the bacteria are dying every minute, how long will it take to have less than 1,000 bacteria? (You are assuming continuous decay)

First, find k using the given conditions:

45% die in 1 minute. That means 55% remain after 1 minute.

\(\displaystyle \L\\0.55(5,000,000)=2,750,000\)

\(\displaystyle \L\\2,750,000=5,000,000e^{k(1)}\)

Solve for k. After you have k, use it to find t(the time for the population to get down to 1000).

\(\displaystyle \L\\1000=5,000,000e^{kt}\).

Solve for t.

 

[sweetliljenny]

thank you so much! i think i arrived at the right answer

0.55 (5,000,000) = 2,750,000
2,750,000 = 5,000,000e ^ (-k)(1)

Divide 5,000,000 on both sides therefore i got:

.55 = e ^ (-k)(1)
ln(.55) = lne ^ -1k

** ln(e^x) = x --> lne ^ -1k = -1k

ln(.55) = -1k
_____ ___
-1 -1

ln(.55) / (-1) = .5978

k = .5978


Q = ae ^ (-k)(t) (For continuous decay --which is why we use -k and not k which is used for continuous growth)

1000 = 5,000,000e ^ (-.5978)(t)

Divide 5,000,000 by both sides =

.0002 = e ^ -.5978t

ln(.0002) = lne ^ -.5978t
ln(.0002) = -.5978t
_______ ______
-.5978 -.5978

ln(.0002) / (-.5978) = approx. 14.2 minutes

t = approx. 14.2 minutes

(i hope i did that correctly )

 

[galactus]

Good. That's what I got.

Except maybe write your answer as \(\displaystyle \L\\t=\frac{ln(5000)}{ln(\frac{20}{11})}\). Just looks better than decimals.

 

[sweetliljenny]

thank you so much galactus you've been a ton of help =). Is there any chance you can help me out with a couple of other problems?

For instance

#1) 10 ^ log(2x+1) = ln2 (Solving the equation exactly for x)

i know the rule: 10 ^ logx = x ..therefore 10 ^ log(2x+1) = 2x + 1

therefore i have 2x + 1 = ln2 ..i just need help from there

#2) ln(x-2) + ln(2x-3) = 2lnx

Rule: Log A x B = Log A + Log B
therefore i have ... ln [ (x-2)(2x-3) ] = lnx ^ 2

** 2lnx is now lnx ^ 2 because of the rule log(b^t) = t x log b **

After factoring i got 2x ^ 2 - 3x - 4x + 6 = x ^ 2

..and i need help after that