Logarithms

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 Problem no. 3

This problem looks complex. But know that logarithm problems can always be very simply resolved if the basic formulae are known well and the advanced formulae shown above are also very clearly formed in your mind.

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.

Harder Problem no. 4

since log b to the base a: we may write it like this:

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

Harder Problem no. 5

Again a complex looking problem. We will sort it out by looking at simple elements inside a complex entity. Then as soon as we find common elements inside the complex entity, we will call that simple entity as ‘a’ or ‘b’ and then do the algebra, (usually it will be a quadratic) and then substitute back to x.
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]

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