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First method draws back
Consider the following equation.
7x = 9
It is a fairly easy equation just won't work since we don't know how to write 9 as a power of 7. Actually if you think regarding it that is exactly what this equation is asking us to determine.
Thus, the method we utilized in the first set of examples won't work. Here the problem is that the x is in the exponent. Due to that all our knowledge regarding solving equations won't do us any good. We require a way to obtain the x out of the exponent and fortunately for us we contain a way to do that.
Recall the following logarithm property from the last section.
log bar = r logb a
Note that to ignore confusion with x's we replaced the x into this property with an a. The significant part of this property is that we can take an exponent &move it into the front of the term.
So, if we had,
logb 7x
We could employ this property as follows.
x logb 7
Now the x in out of the exponent! Certainly now we are stuck along with a logarithm in the difficulty and not only that although we haven't specified the base of the logarithm.
The realism is that we can employ any logarithm to perform this so we have to pick one that we can deal along with. Usually this means that we'll work along with the common logarithm or the natural logarithm.
Example : determine the zeroes of following polynomials. P ( x)= 5x 5 - 20x 4 +5x3 + 50x2 - 20x - 40 = 5 (x + 1) 2 ( x - 2) 3 Solution In this the factoring has been
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