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
thanx for the help though everyone
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 $\displaystyle x= 0$
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.
1. to $\displaystyle (\log_3)^x$: Now we know the base of the logarithmic function but the argument is still missing ...
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 !P.s. do you mind if i PM you with some q's?
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)
$\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:
$\displaystyle \log_3(x) + \log_3(x) = 2\log_3(x)$