# Help with basic log stuff

1. Sep 20, 2006

### preet

Here are some log questions that I don't really have any idea how to do. I'd appreciate any hints/help... (solve for x in all questions)

1. $2^x + 4^x = 8^x$
All I could think of doing was converting them to the same base. I don't see how to simplify the left side of the equation so I can equate the exponents (if that's what I have to do). If I take the log of both sides I just end up getting $x log {2} + x log {4} = log {8}$, and I don't know where to go from there.

2. $log_x {2} = log_2x {8}$
I really have no idea how to approach this.

3. solve for x in terms of y
$y = (e^2x - e^-2x) / 2$
Again, no idea. (specifically, I don't know how to take the natural logarithm of both sides of this equation which is what i think i need to do)

I need some help with this log/ln stuff... if nothing else, are there any good links I can read up on? I forgot pretty much everything I've learned about this stuff.

Thanks
Preet

2. Sep 20, 2006

### Data

For 1, note that you can rewrite it as

$$2^x + (2^x)^2 = (2^x)^3.$$

See if that helps!

I'm going to assume that 2 is actually to solve

$$\log_x2 = \log_{2x}8.$$

Try using $$\log_a b = \frac{\log_c b}{\log_c a}$$ on the RHS to convert all the logs to the same base.

I'm also going to assume 3 is actually to solve

$$y = \frac{e^{2x}-e^{-2x}}{2}$$

for x in terms of y.

Can you solve

$$y = \frac{u + \frac{1}{u}}{2}$$

for u in terms of y? If so, you should be able to do your question too!

Last edited: Sep 20, 2006