OK, I have a good feeling it's going to turn into a complex trinomial.
By substituting with a variable, say "z"
That would give us..
Still not a trinomial..hmm
I see that the factor may be, -1 and 32.
Or, do I somehow break down the 4 into . But still the exponent 2 would not be over the z.
Oh OK. That is the Law of Logarithms for Powers (LLP)
I've used that before. =]
That fraction is REALLY throwing me off..
Next, I think you divide the both sides by log2?
Or do I get the values, like:
It sorta works, and when subbed into the original equation it equals 0.00156
There must be a better way..
Don't let the fraction get to you!
When you take the log of a fraction, the value is negative.
So it is ok to right the solution as . Now by change of base formula , we can rewrite our solution as .
Now when you plug it into verify the solution, we keep in mind that :
.
So it shows that is the solution to our equation.
The problem as given is probably correct, but since you mention trinomials, I have also seen problems where one of the exponents is 2x instead of x, like this:
.
So, if you substitute with another variable (I'll use y), you would get
.
I'd double check wherever you got the problem to make sure that the exponents are correct.
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