Struggling w/ review: finding log values, solving, etc

[holic]

Out of 125 problems, I forgot how to do a few. I was hoping someone who knew the solutions to these problems was browsing =P

1) Given 'log9 7 = 0.8856' and 'log9 4 = 0.6309', find the value of:
a) log9 28
b) log9 49

I tried ".8856 x .6309" for 'a' and got .5587, but the correct answer is 1.5165
I tried ".8856 squared" for 'b' and got .7843, but the correct answer is 1.7712

2) Find the approximate solution: log8(n-3)+log8(n+4)=1

I was trying to isolate the second part... but i got all out of wack.

3) A certain strain of bacteria can grow from 30 to 195 in 5 hours. What is the approximate value of k for the growth formula y=ne^(kt)?

Since y= final amt, n= initial amt, k= constant, and t= time, i got:

195=30e^(5k)
6.5=e^(5k)
log 6.5=5k(log e)

from there I'm lost.

4) Assume $500 is deposited into a savings account. If the interest rate is 8.25% compounded continuously, after how many years will the amount of money in the account be tripled? Use A = Pe^(rt) and round your answer to the nearest 10th.

I tried:

1500 = 500e^((.0825)(2))
3 = e^1.65

probably not starting out right.

Man, these logs are killin me!

Anyways, thanks to all in advanced!

 

[galactus]

Re: Struggling on a few ALG2 Final review problems.

2) Find the approximate solution: log8(n-3)+log8(n+4)=1

I was trying to isolate the second part... but i got all out of wack.

Here's help on one of them:

The trick is to remember your log rules.

\(\displaystyle \L\\log_{8}(n-3)+log_{8}(n+4)=1\)

\(\displaystyle \L\\log_{8}((n-3)(n+4))=1\)

\(\displaystyle \L\\(n-3)(n+4)=8^{1}\)

Now you can solve it?.

 

[soroban]

Re: Struggling on a few ALG2 Final review problems.

Hello, holic!

If you solved the rest of the 125 problems, you're doing fine!

1) Given: \(\displaystyle \,\log_9(7)\,=\,0.8856\,\) and \(\displaystyle \,\log_9(4)\,=\,0.6309\), find the value of:

\(\displaystyle a)\;\log_9(28)\;\;\;\;\;b)\;\log_9(49)\)

You forgot those basic log properties:

\(\displaystyle \;\;\log_b(X\cdot Y)\;=\;\log_b(X)\,+\,\log_b(Y)\;\;\;\;\log_b\left(X^n\right) \;= \;n\cdot\log_b(X)\)


\(\displaystyle (a)\;\log_9(28)\;=\;\log_9(4\cdot7)\;=\;\log_9(4)\,+\,\log_9(7)\)

\(\displaystyle (b)\;\log_9(49)\;=\;\log_9(7^2)\;=\;2\cdot\log_9(7)\)

Got it?

2) Find the approximate solution: \(\displaystyle \,\log_8(n-3)\,+\,\log_8(n+4)\:=\:1\)

Combine: \(\displaystyle \,\log_8[(n-3)(n+4)]\:=\:1\)

Exponentiate: \(\displaystyle \,(n-3)(n+4)\:=\:8^1\;\;\Rightarrow\;\;n^2\,+\,n\,-\,12\:=\:8\)

We have the quadratic: \(\displaystyle \,n^2\,+\,n\,-\,20\:=\:0\)

\(\displaystyle \;\;\)which factors: \(\displaystyle \,(n\,-\,4)(n\,+\,5)\:=\:0\)

\(\displaystyle \;\;\)and has roots: \(\displaystyle \,x\:=\:4,\;-5\)

But \(\displaystyle x\,=\,-5\) is an extraneous root.

Therefore, the only solution is: \(\displaystyle \,x\,=\,4\)

3) A certain strain of bacteria can grow from 30 to 195 in 5 hours.
What is the approximate value of k for the growth formula \(\displaystyle \,y\:=\:ne^{kt}\)?

We are told that, when \(\displaystyle t\,=\,0,\;y\,=\,30\,\) and when \(\displaystyle t\,=\,5,\;y\,=\,195\)

We have two "points" on the function; we can use them to solve for \(\displaystyle n\) and \(\displaystyle k\).

\(\displaystyle (0,\,30):\;\;30\:=\:ne^{k\cdot0}\;\;\Rightarrow\;\;n\,=\,30\)

The function (so far) is: \(\displaystyle \,y\:=\:30e^{kt}\)


\(\displaystyle (5,\,195):\;\;195\:=\:30e^{k\cdot5}\;\;\Rightarrow\;\;6.5\:=\:e^{5k}\;\;\Rightarrow\;\;5k\:=\:\ln(6.5)\)

Therefore: \(\displaystyle \,k\:=\:\frac{1}{5}\ln(6.5)\:\approx\:0.3744\)

4) Assume $500 is deposited into a savings account.
If the interest rate is 8.25% compounded continuously,
after how many years will the amount of money in the account be tripled?
Use \(\displaystyle \,A\:=\e^{rt}\,\) and round your answer to the nearest tenth.

I tried: \(\displaystyle \,1500\:=\:500e^{(0.0825)(2)}\;\) . . . Why "2"?

Do you realize that you have no variables in your equation?
There's nothing to solve for!

You should have: \(\displaystyle \,1500\:=\:500e^{0.0825t}\)

Then: \(\displaystyle \,e^{0.0825t}\:=\:3\;\;\Rightarrow\;\;0.0825t\:=\:\ln(3)\)

Therefore: \(\displaystyle \,t\;=\;\frac{\ln(3)}{0.0825}\;=\;13.31651259\;\approx\;13.3\,\) years

 

[holic]

thank you both so very much.. I had teachers like you guys