Hi
I have come to this stage from solving a logarithm question. Can't seem to proceed. Please help.
6^z+3^z = 2^z
3^z.2^z +3^z = 2^z
3^z = 2^z / (2^z + 2^0)
How can I find the value of z from here?
Thank you.
rgds
gentsl
$\displaystyle 3^{z}(2^{z}+1)=2^{z}$
$\displaystyle 2^{z}+1=(\frac{2}{3})^{z}$
Just looking at this we can see that $\displaystyle \frac{1}{2}+1=\frac{3}{2}$
Which would make z = -1.
You could also make the observation that 1/6+1/3=1/2.
More observation than guess and check.
You could play around and see what you can whip up.
$\displaystyle 6^{z}+3^{z}=2^{z}$
Divide by $\displaystyle 2^{z}$:
$\displaystyle 3^{z}+(\frac{3}{2})^{z}=1$
$\displaystyle 3^{z}+3^{z}e^{-zln(2)}=1$
$\displaystyle 3^{z}(1+e^{-zln(2)})=1$
$\displaystyle 1+e^{-zln(2)}=\frac{1}{3^{z}}$
$\displaystyle 1+2^{-z}=3^{-z}$
Now, since 1+2=3, z must equal -1.
Just playing around, you can always try other things.
Look into the Lambert W function for tricky exponent problems.