1. ## Log problem

The solutions to $(log_3x)^2-3log_3x-4=0$ are $x\:=\:,$

No clue how to work with logs...

2. Originally Posted by davidman
The solutions to $(log_3x)^2-3log_3x-4=0$ are $x\:=\:,$

No clue how to work with logs...
Set u = log_3(x)

$u^2-3u-4=0$

Solve the quadratic for u. Remember only positive values of u will give positive values of x

3. Originally Posted by davidman
The solutions to $(log_3x)^2-3log_3x-4=0$ are $x\:=\:,$

No clue how to work with logs...
To solve this first, take this as a normal quadratic.

If we replace $log_3x$ with $Y$, we get:

$Y^2 - 3Y - 4=0$

Solving this for $Y$ gives you $Y = (-1,4)$.

If we replace $Y$ with $log_3x$, we get the following two equations to solve:

$log_3x = -1$

and

$log_3x = 4$.

You then need to solve these for x.

Have you worked with logs at all? For example, $log_3 81 = 4$, this is the same as $3^4 = 81$. Does this ring any bells?

4. Originally Posted by craig
Have you worked with logs at all? For example, $log_3 81 = 4$, this is the same as $3^4 = 81$. Does this ring any bells?
Nope, completely 100% brand new to logs... but I guess that makes sense.

$log_xM=N \: \:, \: \: x^N=M$

thanks!

$log_3x=-1\:\:,\:\:3^{-1}=x\:\:,\:\:x=\frac{1}{3}$

$log_3x=4\:\:,\:\:3^4=x\:\:,\:\:x=81$

5. Originally Posted by davidman
Nope, completely 100% brand new to logs... but I guess that makes sense.

$log_xM=N \: \:, \: \: x^N=M$

thanks!

$log_3x=-1\:\:,\:\:3^-1=x\:\:,\:\:x=\frac{1}{3}$

$log_3x=4\:\:,\:\:3^4=x\:\:,\:\:x=81$
Yep