1 Word problem and 1 function problem

• Apr 19th 2010, 04:21 PM
murkr
1 Word problem and 1 function problem
you guys are all very good at math, i love these forums. I understand more things here then i do most the time in class, i have a test coming up and i really need help understanding these 2 problems, can someone please solve these problems and explain the steps, i would really appreciate it.

i got a review packet, i scanned the problems because i dont know how to type them
http://img694.imageshack.us/img694/9...wrk2222222.jpg

problem #28 and #29 are the ones im having problem with and need the help.

also was i correct with #27? was it A? thanks everyone
• Apr 19th 2010, 04:42 PM
pickslides
28) For $\displaystyle 2^x = 15$

consider the rule when $\displaystyle a^b=c \implies b = \log_ac$ this will give you the answer

29) And for $\displaystyle A= 456$ in $\displaystyle A= 282 e^{0.04t}$ then

$\displaystyle 456= 282 e^{0.04t}$ solve for $\displaystyle t$
• Apr 19th 2010, 05:03 PM
Krahl

on both sides of the equation use the following
$\displaystyle a^x=b^y \Longrightarrow \ln (a^x)=\ln(b^y) \Longrightarrow x\ln(a)=y\ln(b)$

you may need to use

$\displaystyle \log_ac=\frac{\log_{10}c}{\log_{10}a}$ if you need to change base to use your calculator.

Edit: My bad you don't need to evaluate $\displaystyle \log_ac$ but worth knowing anyway :s
• Apr 20th 2010, 06:02 PM
murkr
Quote:

Originally Posted by Krahl

on both sides of the equation use the following
$\displaystyle a^x=b^y \Longrightarrow \ln (a^x)=\ln(b^y) \Longrightarrow x\ln(a)=y\ln(b)$

you may need to use

$\displaystyle \log_ac=\frac{\log_{10}c}{\log_{10}a}$ if you need to change base to use your calculator.

Edit: My bad you don't need to evaluate $\displaystyle \log_ac$ but worth knowing anyway :s

i have spent this last hour trying to figure out #27. can you please solve this problem, is it B or C?
• Apr 20th 2010, 08:03 PM
sa-ri-ga-ma
Quote:

Originally Posted by murkr
i have spent this last hour trying to figure out #27. can you please solve this problem, is it B or C?

For #27
xln(6/7) = (1-x)ln(2)
Now solve for x.
• Apr 20th 2010, 09:17 PM
murkr
Quote:

Originally Posted by sa-ri-ga-ma
For #27
xln(6/7) = (1-x)ln(2)
Now solve for x.

yes i got to the next step

x * (ln(6/7) + ln[2]) = ln[2]

can you please work the question out in steps, i understand it a lot more when i see it done in steps, thanks for all your help. and actually trying to get me to figure it out myself and not just give me the answer haha. but im really having a hard time and my test is tomrrow so please solve this in steps for me, thank you.
• Apr 20th 2010, 10:59 PM
sa-ri-ga-ma
x * (ln(6/7) + ln[2]) = ln[2]

One last step for the result.
If ax= b, then
x = .....?
• Apr 20th 2010, 11:21 PM
murkr
Quote:

Originally Posted by sa-ri-ga-ma
x * (ln(6/7) + ln[2]) = ln[2]

One last step for the result.
If ax= b, then
x = .....?

is it C?? thats what i got, if its wrong im stumped please finish.. i just forgot how to do this
• Apr 21st 2010, 08:01 AM
Krahl
$\displaystyle \frac{6}{7}^x= (2)^{1-x}$

take log of both sides

$\displaystyle ln\frac{6}{7}^x= ln(2)^{1-x}$

the exponent comes down

$\displaystyle xln\frac{6}{7}= (1-x)ln(2)$

Collect like terms

$\displaystyle xln\frac{6}{7}= 1ln(2)-xln(2)$

$\displaystyle xln\frac{6}{7}+xln(2)= 1ln(2)$

$\displaystyle x(ln\frac{6}{7}+ln(2))= 1ln(2)$

and divide both sides by $\displaystyle ln\frac{6}{7}+ln(2))$ and yes you get C