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Solving logarithmic equations

Q:  Solve
log16x + log4x + log2x = 7

A:  It is easier (though not always necessary) to have all of the logs in the same base to proceed.  Since that is not the case, we need to use the change of base formula to put all of the logs into the same base.

Change of base formula says:
logab = logcb / logcwhere c can be any base of your choosing

What base should we go to?  Since 16, 4 and 2 can all be formed by powers of 2, let’s go to base 2:

log16x + log4x + log2x = 7

log2x / log216 + log2x / log24 + log2x / log22 = 7

Now, we can simplify:

log2x / 4 + log2x / 2 + log2x / 1 = 7

1/4 log2x + 1/2 log2x + log2x = 7

7/4 log2x = 7

log2x = 7 *(4/7)

log2x = 4

(remember, this reads:  the power you put on 2 to get x is 4)

x = 24

x = 16

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Half-Life Word Problem

Q:  A scientist has 256 g of goo. After 195 minutes, her sample has decayed to 16 g. What’s the half-life of the goo in minutes? (Assume exponential decay)

Answer:

First we need to set up the problem with our decay formula.  Depending on your school / teacher / book, the variables in the formulas may be different letters, but they all mean the same thing:

P = A*e^(rt)

P is the ending amount, A is the initial amount, r is the growth rate/decay, and t is time.

So, the set-up with our info:

16 = 256*e^(r*195)

Now we need to solve for r:

16 = 256*e^(r*195)

1/16 = e^(r*195)

Take the natural log of both sides: Continue reading Half-Life Word Problem