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## #1 2005-11-08 02:11:32

Zaara
Guest

### logarithms

Hey there,

I swear, my brain is not working at the moment and I have a big test tomorrow.

I would be forever thankful if someone could help me with these problems ASAP

solve: log27 X = 1 - log27 (x-0.4)
(27 is the base)

and solve: log9 81 + log9 (1/9) + log9 3 = log9 X
(9 is the base)

Thank you!

## #2 2005-11-08 04:22:35

mathsyperson
Moderator

Offline

### Re: logarithms

The second one is easy because you can work out the value of each term without involving logs.

9^2 = 81, so log9 81 = 2; 9^-1 = 1/9, so log9 (1/9) = -1 and so on.

This makes the equation 2 - 1 + 1/2 = log9 X, or 1.5 = log9 X.
Putting each term as a power of 9 gives 9^1.5 = X and so X = 27.

The first is trickier because you can't convert out of logs like that.
One of the laws of logs is that log a - log b = log (a/b), and to apply that rule we need to turn 1 into a log.
27^1 = 27, so 1 = log27 27.

So now the equation is log27 X = log27 27 - log27 (X-0.4) and applying the above rule gives log27 X = log27 (27/(X-0.4)).

As both sides are now in terms of logs, we can remove them to get X = 27/(X-0.4) and from there it's just a quadratic equation.

X(X-0.4) = 27
X² - 0.4X - 27 = 0

Using the Quadratic Equation Solver gives solutions of X = 5.4 and -5, but you cannot have negative logs so the -5 is disregarded. X = 5.4.

Edit: I got bored, so decided to turn the word 'log' brown. Sorry.

Why did the vector cross the road?
It wanted to be normal.

## #3 2005-11-08 04:58:38

Zaara
Guest

### Re: logarithms

My savior.
My hero / heroine.

lol, in other words, THANK YOU SO MUCH