What is the value of x/y -1 if 4^x=8^y. . .? Plz help.

Question

lgbasallote · Answer

4^x = (2^2)^x
8^y = (2^3)^y

2^2x = 2^3y
2x = 3y via rule of same base exponential properties

x = 3y/2

substitute...

3y/2/y - 1
3/2 - 1 = 1/2

anonymous · Answer

$$\LARGE 4^x=8^y\quad \Longrightarrow \log(4)^x=\log(8)^y$$
$$\LARGE x\cdot \log(4)=y\cdot \log(8) $$
$$\LARGE \frac xy =\frac{\log(8)}{\log(4)} $$
$$\LARGE \frac xy =\frac{\log(2)^3}{\log(2)^2} $$
$$\LARGE \frac xy =\log_{2^2}(2)^3 $$
$$\LARGE \frac xy =\frac12 \cdot 3\cdot \log_{2}(2) $$
$$\LARGE \frac xy =\frac32 \cdot 1 $$
... Now? :)

lgbasallote · Answer

$\large \mathsf{\text{interesting method kreshnik /:)}}$

anonymous · Answer

I love "log"  I hate "ln"    LOL

lgbasallote · Answer

lol you couldve just used the same base technique :P

anonymous · Answer

I know, anyway Logs are better when there's no way to get same bases :F (Actually I didn't even check for it, I just used Log immediately !) LOL

lgbasallote · Answer

lol  =)) you think too complex :P