# Logarithmic Functions....help =)

• Jan 25th 2009, 11:19 PM
Baginoman
Logarithmic Functions....help =)
Hey guys, having a little trouple with functions. Heres the problems:

1) 4^x+2-4^x=15

2)7^x^2=10

3) 2^x=3^x

and i have no idea how to post other q's with my keyboard(Headbang)

thanx for the help though everyone(Happy)
• Jan 25th 2009, 11:59 PM
earboth
Quote:

Originally Posted by Baginoman
Hey guys, having a little trouple with functions. Heres the problems:

1) 4^x+2-4^x=15

2)7^x^2=10

3) 2^x=3^x

and i have no idea how to post other q's with my keyboard(Headbang)

thanx for the help though everyone(Happy)

1. I assume that you wanted to solve these equations(?)

2. I assume that you meant:

$\displaystyle 4^{x+2}-4^x=15~\implies~16 \cdot 4^x-4^x=15~\implies~ 15\cdot 4^x=15$

I leave the rest for you.

$\displaystyle 7^{x^2} = 10~\implies~x^2=\log_7(10)~\implies~x=\sqrt{\log_7 (10)}$

$\displaystyle 2^x=3^x~\implies~1=\dfrac{3^x}{2^x} = \left(\dfrac32\right)^x$

Now use the definiton $\displaystyle a^0=1\ ,\ a > 0$

to get $\displaystyle x= 0$
• Jan 26th 2009, 04:17 PM
Baginoman
awesome, those confused me lol

1) log3^x=log3(1/x)+4 (keep in mind both 3's are log of 3)

2) log(7y + 1)= 2 log(y + 3) - log 2

3) log(3x - 4)/log x =2

thank you again guys
• Jan 26th 2009, 05:24 PM
becker89
second question: just use logarithm rules,
the 2 goes back to the exponent,then you have a difference of logarithms
third question: change of bases rule
• Jan 26th 2009, 08:39 PM
earboth
Quote:

Originally Posted by Baginoman
...

1) log3^x=log3(1/x)+4 (keep in mind both 3's are log of 3)

2) log(7y + 1)= 2 log(y + 3) - log 2

3) log(3x - 4)/log x =2

...

1. Do us and do yourself a favour and start a new thread if you have new questions. Otherwise you risk that nobody will notice your need for help.

2. I don't understand the equation in 1). Do you mean:

$\displaystyle (\log(3))^x$ or $\displaystyle (\log_3)^x$ or $\displaystyle \log(3^x)$ ...

3. If you have some difficulties to use becker89 hint, here is a starter:

$\displaystyle \log(7y+1)=2\log(y+3)-\log(2)~\implies~$ $\displaystyle \log(7y+1)+\log(2)=2\log(y+3)~\implies~\log(14y+2) =\log((y+3)^2)$

Now de-logarithmize. You'll get a quadratic equation in y. My result is: y = 7 or y = 1

4. Multiply the equation through by log(x):

$\displaystyle \log(3x-4)=2\log(x)~\implies~\log(3x-4)=\log(x^2)$

De-logarithmize and solve the quadratic equation. This equation doesn't have a real solution.
• Jan 26th 2009, 08:54 PM
Baginoman
Quote:

Originally Posted by earboth
1. Do us and do yourself a favour and start a new thread if you have new questions. Otherwise you risk that nobody will notice your need for help.

2. I don't understand the equation in 1). Do you mean:

$\displaystyle (\log(3))^x$ or $\displaystyle (\log_3)^x$ or $\displaystyle \log(3^x)$ ...

3. If you have some difficulties to use becker89 hint, here is a starter:

$\displaystyle \log(7y+1)=2\log(y+3)-\log(2)~\implies~$ $\displaystyle \log(7y+1)+\log(2)=2\log(y+3)~\implies~\log(14y+2) =\log((y+3)^2)$

Now de-logarithmize. You'll get a quadratic equation in y. My result is: y = 7 or y = 1

4. Multiply the equation through by log(x):

$\displaystyle \log(3x-4)=2\log(x)~\implies~\log(3x-4)=\log(x^2)$

De-logarithmize and solve the quadratic equation. This equation doesn't have a real solution.

sorry about that, and yes the first one is $\displaystyle (\log_3)^x$

also if its not a hassle, how does logX^2=2 come out as (10, -10)??

P.s. do you mind if i PM you with some q's?
• Jan 26th 2009, 09:13 PM
earboth
Quote:

Originally Posted by Baginoman
sorry about that, and yes the first one is $\displaystyle (\log_3)^x$

also if its not a hassle, how does logX^2=2 come out as (10, -10)??

P.s. do you mind if i PM you with some q's?

1. to $\displaystyle (\log_3)^x$: Now we know the base of the logarithmic function but the argument is still missing ...(Wondering)

2. If $\displaystyle \log$ means $\displaystyle \log_{10}$ the the last equation can be solved:

$\displaystyle \log(x^2)=2~\implies~10^{\log(x^2)} = 10^2~\implies~x^2=100$

Quote:

P.s. do you mind if i PM you with some q's?
Don't do that: Give other members of MHF a chance :D!
• Jan 26th 2009, 09:20 PM
Baginoman
Quote:

Originally Posted by earboth
1. to $\displaystyle (\log_3)^x$: Now we know the base of the logarithmic function but the argument is still missing ...(Wondering)

2. If $\displaystyle \log$ means $\displaystyle \log_{10}$ the the last equation can be solved:

$\displaystyle \log(x^2)=2~\implies~10^{\log(x^2)} = 10^2~\implies~x^2=100$

Don't do that: Give other members of MHF a chance :D!

hahah thankyou very much, your talent makes me want to learn math to a greater extent!

Plus i would post another thread for more help, but i kinda need an answer within the next half hour or so (Rofl)
• Jan 26th 2009, 10:30 PM
earboth
Quote:

Originally Posted by Baginoman
...
1) log3^x=log3(1/x)+4 (keep in mind both 3's are log of 3)

...

I believe that I found the solution to this cryptic equation:

Probably you mean:

$\displaystyle \log_3(x)=\log_3\left(\frac1x \right) + 4~\implies~\log_3(x)=-\log_3\left(x \right) + 4 ~\implies~$ $\displaystyle 2\log_3(x)=4~\implies~x^2=81$

Since x > 0 there is only one solution. (x = 9)
• Jan 26th 2009, 10:38 PM
Baginoman
Quote:

Originally Posted by earboth
I believe that I found the solution to this cryptic equation:

Probably you mean:

$\displaystyle \log_3(x)=\log_3\left(\frac1x \right) + 4~\implies~\log_3(x)=-\log_3\left(x \right) + 4 ~\implies~$ $\displaystyle 2\log_3(x)=4~\implies~x^2=81$

Since x > 0 there is only one solution. (x = 9)

yep thats its lol. hmmm where did the 2log come from?
• Jan 26th 2009, 10:49 PM
earboth
Quote:

Originally Posted by Baginoman
yep thats its lol. hmmm where did the 2log come from?

$\displaystyle \log_3(x)=\log_3\left(\frac1x \right) + 4~\implies~\log_3(x)=-\log_3\left(x \right) + 4 ~\implies~$

To get rid of the log at the RHS you have to add $\displaystyle \log_3(x)$ on both sides of the equation. And then you'll get at the LHS:

$\displaystyle \log_3(x) + \log_3(x) = 2\log_3(x)$