ARITHMETIC ALGEBRA GEOMETRY QUADRATIC function STRAIGHT LINES
The basics of logarithm ought to be known to the reader. Here we have shown a few of the slightly advanced formulae for logarithms.
Logarithm is a lot of fun. It has many colourful properties. Become friends with them and you two will do famously together 🙂
HARDER PROBLEMS SOLVED
Harder Problem no. 1
Harder Problem no. 2
Here, we have taken up the elements one by one, step by step.
If you follow a systematic process, you can eat an elephant bit by bit 🙂
In order to solve the problem, we have taken log of both sides.
Remember to convert decimal into a fraction if need be, as here.
Another useful idea is to take a K for any repeated element inside a complex expression, like we have done here. Then dealing with K is much simpler. Later convert back into x.
Harder Problem no. 3
As shown in the previous problem, we have taken a variable ‘a’ to simplify the equation and end up taking another variable “b” to arrive at a quadratic equation which we solve and then substitute back to x when we get our solution.
We have covered Quadratic extensively here.
log (a) b
imagine the number in bracket as the base.
Now this can be separated into log b / log a and that makes it very simple.
Besides, a log p = log p^a
is also very useful and we will use this extensively and interchangeably
In our computation of the RHS (Right Hand Side), we have jumped a few steps e.g.
(7^log A)(7^log B) = 7^ (log A + log B) and log A + log B = log AB
In this problem, (2/3)^(1/x) is taken as ‘b’, as shown.
x+1/x or x-1/x are very useful algebraic entities and has interesting properties of its square or cube. After a further substitution bringing in a new variable K, we end up with a cubic equation. For any polynomial, a shortcut to finding the roots lies in noticing the constant term in the polynomial (in the above problem it is 3 which means that the roots will be factors of 3. So, we tried to put K=1 in the cubic equation and found it to be satisfying the cubit, thus K-1 would be a factor. Writing down the factor K-1, we have computed the other factor which is a quadratic. Immediately, we note the Discriminant, D of the Quadratic and find that D<0 i.e. D is negative and as we know from basic analysis of the quadratic function, there will be no root. So, we have only one factor K-1 => K = 1. Going back in our substitution chain, we come to ‘b’ which again gives us a quadratic in terms of b and thus we arrive at the two roots being the values of b, one negative and one positive value of b. But b is n^(1/x) and this cannot be negative since in exponent, we take only positive value of the base. Besides the negative value of the exponent translates into the reciprocal and thus b cannot be negative. This eliminates one value of b and we end up with the value of x [which we have put in a blue box as shown in the diagram]