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

No clue how to work with logs...

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- Jan 24th 2010, 07:59 AM #1

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- Jan 24th 2010, 08:01 AM #2

- Jan 24th 2010, 08:12 AM #3
To solve this first, take this as a normal quadratic.

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

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

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

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

$\displaystyle log_3x = -1$

and

$\displaystyle log_3x = 4$.

You then need to solve these for x.

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

- Jan 24th 2010, 07:40 PM #4

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- Jan 24th 2010, 10:42 PM #5