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Expert: - 6/22/2009


Question
1. log3x-log3(x-3)=2
2.log2x-log2(x-2)=3
3. 2^x=6
4. e^x+3=8
5.log13x/5
6. log381=x

Answer
1. It is known that with log, subtracting two is the same as dividing the insides.  In other words,
log x - log x-3 = log (x/(x-3)), all in base 3.

For this to be equal to 2, take both sides as exponets on 3.
Since taking an exponet is the inverse of a log,
on the left we just get x/(x-3).

On the right we get 3^2 = 9.

This tells us the x/(x-3) = 9, or x = 9(x-3), which goes to
x = 9x - 27, so -8x = 27, or x = 27/8 = 3 3/8.


2. We know that log 2x - log x-2 = log (x/(x-2)).
Since the base is 2, take both sides as exponets of 2.
This makes x/(x-2) = 2^3 = 8.

From there, since x/(x-2) = 8, we can also say that x = 8(x-2).
Multiplying it out gives x = 8x - 16, or -7x = 16, or x = -16/7.
This can also be x = -2 2/7.


3. We are given 2^x = 6.  Take log of both sides, base 2.
These cancel on the left, so we now have x = log 6, base 2.
We both know that 6 = 2 * 3, so log 6 = log 2 + log 3 = 1 + log 3.
This gives x = 1 + log 3.


4. We are given e^x + 3 = 8.  
Subtracting 3 gives from both sides gives us e^x = 5.
Taking ln of both sides gives us
x = ln 5.

5. What is log 13x/5 ?

If that is the problem,
I could say that is was log 13 + log x - log 5 =
log 13/5 + log x.

Is that suppose to be log 13x = 5?
In that case, if the log is base 10, 13x = 10^5 = 100,000.
That means x = 100,000/13.


6. log 381 = x.  I'm not sure what to do.

The variable x is already solved for and can be found on a calculator or in Excel.

I can say that 381 = 10^x, but that doesn't do much.

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