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

2. $\displaystyle 3^{z}(2^{z}+1)=2^{z}$

$\displaystyle 2^{z}+1=(\frac{2}{3})^{z}$

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.

3. Thanks Galactus.

That was done through observation (guess and check), is there any other way to do it?

gentsl

4. 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.